2 Marks Questions - Chemistry STD 11 Science Questions - Vidyadip

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2 Marks Questions

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Question 12 Marks

Use the periodic table to answer the following questions.
Identify an element that would tend to lose two electrons.

Answer

An element having two valence electrons will lose two electrons easily to attain the stable noble gas configuration. The general electronic configuration of such an element will be $\mathrm{ns}^2$.
This is the electronic configuration of group 2 elements. The elements present in group 2 are $\mathrm{Be}, \mathrm{Mg}$, $\mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}$.

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Question 22 Marks

What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
$\mathrm{Mg}^{2+}$

Answer

$\mathrm{Mg}^{2+}$ ion has $12-2=10$ electrons. Thus, the species isoelectronic with it will also have 10 electrons. Some of its isoelectronic species are $\mathrm{F}^{-}$ion ( $9+1=10$ electrons), Ne ( 10 electrons), $\mathrm{O}^{2-}$ ion ( $8+2$ $=10$ electrons), and $\mathrm{Al}^{3+}$ ion ( $13-3=10$ electrons).

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Question 32 Marks

What is the basic difference between the terms electron gain enthalpy and electronegativity?

Answer

Electron gain enthalpy is the measure of the tendency of an isolated gaseous atom to accept an electron, whereas electronegativity is the measure of the tendency of an atom in a chemical compound to attract a shared pair of electrons.

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Question 42 Marks

Assign the position of the element having outer electronic configuration.
$(n-1) d^2 n s^2$ for $n=4$,

Answer

Since $\mathrm{n}=4$, the element belongs to the $4^{\text {th }}$ period. It is a d-block element as d -orbital's are incompletely filled.
There are 2 electrons in the d-orbital.
Thus, the corresponding group of the element
$=$ Number of s-block groups + number of d-block groups
$=2+2$
$=4$
Therefore, it is a $4^{\text {th }}$ period and $4^{\text {th }}$ group element. Hence, the element is Titanium.

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Question 52 Marks

What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
$\mathrm{Rb}^{+}$

Answer

$\mathrm{Rb}^{+}$ion has $37-1=36$ electrons. Thus, the species isoelectronic with it will also have 36 electrons. Some of its isoelectronic species are $\mathrm{Br}^{-}$ion ( $35+1=36$ electrons), Kr ( 36 electrons), and $\mathrm{Sr}^{2+}$ ion $(38-2=36$ electrons).

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Question 62 Marks

The first ($(\Delta_{1}\text{H}_{1})$ and the second $(\Delta_{1}\text{H}_{1})$ ionization enthalpies (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) and the $(\Delta_{1}\text{H}_{1})$ electron gain enthalpy (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) of a few elements are given below:

Elements	$\Delta\text{H}_{1}$	$\Delta{\text{H}}_{2}$	$\Delta_{\text{eg}}\text{H}$
I	520	7300	-60
II	419	3051	-48
III	1681	3374	-328
IV	1008	1846	-295
V	2372	5251	+48
VI	738	1451	-40

Which of the above elements is likely to be:
The most reactive non-metal.

Answer

III: The element III has very high negative electron gain enthalpy hence, it is most reactive non-metal.

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Question 72 Marks

Assign the position of the element having outer electronic configuration. $(n-2) f^7(n-1) d^1 n s^2$ for $n=6$, in the periodic table.

Answer

Since $\mathrm{n}=6$, the element is present in the $6^{\text {th }}$ period. It is an $f$ - block element as the last electron occupies the f-orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe] $4 f^7 ~5 d^1 ~6 s^2$. Thus, its atomic number is $54+7+2+1=$ 64. Hence, the element is Gadolinium.

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Question 82 Marks

Describe the theory associated with the radius of an atom as it.
Gains an electron.

Answer

The distance between the nucleus and the outermost shell of an ion is known as ionic radius.
The gain of an electron leads to the formation of an anion. The radius of the anion is larger than the atomic radius of its parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge which acts on more electrons so, each electron is held less tightly and thereby the electron cloud expands.

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Question 92 Marks

The first $(\Delta_{1}\text{H}_{1})$ and the second $(\Delta_{1}\text{H}_{1})$ ionization enthalpies (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) and the $(\Delta_{1}\text{H}_{1})$ electron gain enthalpy (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) of a few elements are given below:

Elements	$\Delta\text{H}_{1}$	$\Delta{\text{H}}_{2}$	$\Delta_{\text{eg}}\text{H}$
I	520	7300	-60
II	419	3051	-48
III	1681	3374	-328
IV	1008	1846	-295
V	2372	5251	+48
VI	738	1451	-40

Which of the above elements is likely to be:
The least reactive non-metal.

Answer

IV : Element IV has high negative electron gain enthalpy but ionization energy is not that high hence, it is least reactive non-metal.

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Question 102 Marks

Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.

Answer

The ionization enthalpy of an atom depends on the number of electrons and protons (nuclear charge) of that atom. Now, the isotopes of an element have the same number of protons and electrons. Therefore, the first ionization enthalpy for two isotopes of the same element should be the same.

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Question 112 Marks

Assign the position of the element having outer electronic configuration.
$n s^2 n p^4$ for $n=3$

Answer

Since $\mathrm{n}=3$, the element belongs to the $3^{\text {rd }}$ period. It is a p -block element since the last electron occupies the p-orbital.
There are four electrons in the p-orbital. Thus, the corresponding group of the element
$=\text { Number of s-block groups }+ \text { number of d-block groups }+ \text { number of p-electrons }$
$=2+10+4$
$=16$
Therefore, the element belongs to the $3^{\text {rd }}$ period and $16^{\text {th }}$ group of the periodic table. Hence, the element is Sulphur.

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Question 122 Marks

Describe the theory associated with the radius of an atom as it.
loses an electron.

Answer

The distance between the nucleus and the outermost shell of an ion is known as ionic radius.
The removal of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same and since it is now acting on lesser number of electrons and pulls them closer, the ion is smaller.

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Question 132 Marks

Use the periodic table to answer the following questions.
Identify an element that would tend to gain two electrons.

Answer

An element is likely to gain two electrons if it needs only two electrons to attain the stable noble gas configuration. Thus, the general electronic configuration of such an element should be $\mathrm{ns}^2 \mathrm{~np}^4$. This is the electronic configuration of the oxygen family.

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Question 142 Marks

What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
$\text{F}^{-}$

Answer

$\mathrm{F}^{-}$ion has $9+1=10$ electrons. Thus, the species isoelectronic with it will also have 10 electrons. Some of its isoelectronic species are $\mathrm{Na}^{+}$ion ($11 - 1=10$ electrons), Ne (10 electrons), $\mathrm{O}^{2-}$ ion ( $8+$ $2=10$ electrons), and $\mathrm{Al}^{3+}$ ion ( $13-3=10$ electrons).

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Question 152 Marks

The first $(\Delta_{1}\text{H}_{1})$ and the second $(\Delta_{1}\text{H}_{1})$ ionization enthalpies (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) and the $(\Delta_{1}\text{H}_{1})$ electron gain enthalpy (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) of a few elements are given below:

Elements	$\Delta\text{H}_{1}$	$\Delta{\text{H}}_{2}$	$\Delta_{\text{eg}}\text{H}$
I	520	7300	-60
II	419	3051	-48
III	1681	3374	-328
IV	1008	1846	-295
V	2372	5251	+48
VI	738	1451	-40

Which of the above elements is likely to be:
The most reactive metal.

Answer

II : The element II has the least first ionization enthalpy hence, it is most reactive metal.

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Question 162 Marks

Why do elements in the same group have similar physical and chemical properties?

Answer

The physical and chemical properties of elements depend on the number of valence electrons. Elements present in the same group have the same number of valence electrons. Therefore, elements present in the same group have similar physical and chemical properties.

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Question 172 Marks

What is the basic theme of organisation in the periodic table?

Answer

The basic theme of organization of elements in the periodic table is to simplify and systematize the study of the properties of all the elements and millions of their compounds. This has made the study simple because the properties of elements are now studied in form of groups rather than individually.

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Question 182 Marks

The first $(\Delta_{1}\text{H}_{1})$ and the second $(\Delta_{1}\text{H}_{1})$ionization enthalpies (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) and the $(\Delta_{1}\text{H}_{1})$ electron gain enthalpy (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) of a few elements are given below:

Elements	$\Delta\text{H}_{1}$	$\Delta{\text{H}}_{2}$	$\Delta_{\text{eg}}\text{H}$
I	520	7300	-60
II	419	3051	-48
III	1681	3374	-328
IV	1008	1846	-295
V	2372	5251	+48
VI	738	1451	-40

Which of the above elements is likely to be:
The metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?

Answer

I : The element 1 has very low value of first ionization energy but very high second ionization energy. Hence, it will form a stable covalent I halide of the formula MX.

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Question 192 Marks

Use the periodic table to answer the following questions.
Identify an element with five electrons in the outer subshell.

Answer

The electronic configuration of an element having 5 electrons in its outermost subshell should be $\mathrm{ns}^2$ $n p^5$. This is the electronic configuration of the halogen group.
Thus, the element can be $\mathrm{F}, \mathrm{CL}, \mathrm{Br}, \mathrm{I}$, or At.

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Question 202 Marks

Which of the elements Na, Mg, Si and P would have the greatest difference between the first and second ionisation enthalpies? Briefly explain your answer.

Answer

Na has greatest difference between first and second ionisation enthalpies because Na has stable electronic configuration, i.e., $1 s^2 ~2 s^2 ~2 p^6$, therefore, it has very high second ionisation energy.

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Question 212 Marks

Explain, why the electronegativity values of noble gases are zero while those of halogens are the highest in each period?

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Question 222 Marks

Predict the periods and blocks to which each of the following elements belong?
a. ${ }_{13} \mathrm{Al}$
b. ${ }_{24} \mathrm{Cr}$
c. ${ }_{29} \mathrm{Cu}$
d. ${ }_{11} \mathrm{Na}$

Answer

a. ${ }_{13} \mathrm{Al}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}^1$
[Third period and p-block]
b. ${ }_{24} \mathrm{Cr}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 4 s^1, 3 d^5$
[Fourth period and d-block]
c. ${ }_{29} \mathrm{Cu}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 4 \mathrm{~s}^1, 3 \mathrm{~d}^{10}$
[Fourth period and d-block]
d. ${ }_{11} \mathrm{Na}=1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^1$
[Third period and s-block]

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Question 232 Marks

Which out of Cu, Fe, V, Ag is not in first transition series (3d series).
Which element in Group 13 forms most acidic oxide? Give its formula.

Answer

Ag belongs to 4d series (2nd series) others belong to 3d series.
Boron forms acidic oxide. The formula is $\mathrm{B}_2 \mathrm{O}_3$.

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Question 242 Marks

The first $(\Delta_{1}\text{H}_{1})$ and the second $(\Delta_{1}\text{H}_{1})$ ionization enthalpies (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) and the $(\Delta_{1}\text{H}_{1})$ electron gain enthalpy (in $\mathrm{kJ} \mathrm{~mol}^{-1}$ ) of a few elements are given below:

Elements	$\Delta\text{H}_{1}$	$\Delta{\text{H}}_{2}$	$\Delta_{\text{eg}}\text{H}$
I	520	7300	-60
II	419	3051	-48
III	1681	3374	-328
IV	1008	1846	-295
V	2372	5251	+48
VI	738	1451	-40

Which of the above elements is likely to be:
The metal which can form a stable binary halide of the formula $\mathrm{MX}_2(\mathrm{X}=$ halogen $)$.

Answer

VI : The first and second ionization energies of element VI indicate that it can form a stable binary halide.

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Question 252 Marks

Name the species that will be isoelectronic with the following atoms or ions:
i. $Na$
ii. $\mathrm{Cl}^{-}$
iii. $\mathrm{Ca}^{2+}$
iv. $\mathrm{Rb}^{+}$

Answer

Isoelectronic species are those which have same number of electrons.
i. Na has 10 electrons. Therefore, the species $\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Mg}^{2+}, \mathrm{Al}^{3+}$ etc., each of which has also 10 electrons and hence, isoelectronic with it.
ii. $\mathrm{Cl}^{-}$has 18 electrons. Therefore, the species $\mathrm{P}^{3-}, \mathrm{S}^{2-}, \mathrm{Ar}, \mathrm{K}^{+}$and $\mathrm{Ca}^{2+}$, each one of which contains 18 electrons and hence, isoelectronic with it.
iii. $\mathrm{Ca}^{2+}$ has 18 electrons. Therefore, the species $\mathrm{P}^{3-}, \mathrm{S}^{2-}, \mathrm{Ar}$ and $\mathrm{K}^{+}$, each of which also contains 18 electrons and hence, isoelectronic with it.
iv. $\mathrm{Rb}^{+}$has 36 electrons. Therefore, the species $\mathrm{Br}^{-}, \mathrm{Kr}$ or $\mathrm{Sr}^{2+}$ each of which also has 36 electrons and hence, isoelectronic with it.

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Question 262 Marks

For each of the following pairs, predict which one has lower first ionisation enthalpy?
i. N or O
ii. Na or $\mathrm{Na}^{+}$
iii. $\mathrm{Be}^{+}$or $\mathrm{Mg}^{2+}$
iv. I or $I^{-}$

Answer

i. O has lower $\Delta_i \mathrm{H}_1$ than N because in case of $\mathrm{O}\left(1 s^2 2 s^2 2 p^4\right)$, loss of an electron gives $\mathrm{O}^{+}$which has stable halffilled electronic configuration while in case of $\mathrm{N}\left(1 s^2 2 s^2 2 p^3\right)$, an electron has to be removed from stable halffilled electronic configuration.
ii. Na has lower $\Delta_{\mathrm{i}} \mathrm{H}_1$ than $\mathrm{Na}^{+}$because of the following two reasons.
- In case of $\mathrm{Na}^{+}\left(1 s^2 2 s^2 2 p^6\right)$, an electron has to be lost from a stable inert gas configuration but in case of $\mathrm{Na}\left(1 s^2 2 s^2 2 p^6 3 s^1\right)$, loss of an electron gives stable inert gas configuration.
- Na is a neutral atom but $\mathrm{Na}^{+}$is positively charged ion.
iii. $\mathrm{Be}^{+}$has lower $\Delta_{\mathrm{i}} \mathrm{H}_1$ than $\mathrm{Mg}^{2+}$ because in case of $\mathrm{Be}^{+}$, the removal of one electron gives a stable inert gas to configuration but in case of $\mathrm{Mg}^{2+}$, the electron has to be removed from the stable inert gas configuration. iv. I has lower $\Delta_i \mathrm{H}_1$ than $\mathrm{I}^{-}$because in case of $\mathrm{I}^{-}\left(5 \mathrm{~s}^2 5 \mathrm{p}^5\right)$, an electron has to be removed from a stable inert gas configuration.

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Question 272 Marks

Name a species that will be isoelectronic with each of the following atoms or ions:
i. $\mathrm{Ne}$
ii. $\mathrm{Cl}^{-}$
iii. $\mathrm{Ca}^{2+}$
iv. $\mathrm{Rb}$

Answer

i. $\mathrm{Na}^{+}$
ii. $Ar$
iii. $S^{2-}$
iv. $Y^{2+}$

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Question 282 Marks

i. Which is largest in size $\mathrm{Cu}^{+}, \mathrm{Cu}^{2+}$ or Cu , and why?
ii. Which element in periodic table has highest I.E. (ionisation energy)?
iii. Which element is more metallic-Mg or Al and why?

Answer

Cu is largest due to less effective nuclear charge. It has 29 electrons, 29 protons, $\mathrm{Cu}^{+}$has 28 electrons and 29 protons, $\mathrm{Cu}^{2+}$ has 27 electrons and 29 protons.
He has highest ionisation energy because its outermost shell is completely full and most stable.
Mg is more metallic due to lower ionisation energy.

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Question 292 Marks

Arrange in increasing order of:

Atomic size I, F, Cl, Br.
Oxidising power I, F, Cl, Br.

Answer

F < Cl < Br < I due to increase in number of shells.
I < Br < Cl < F due to ability to gain electrons increases with decrease in size.

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Question 302 Marks

How does basic character of oxides and hydroxides down the group in alkali metals change? Why?
How does reducing power of elements vary in Group I?

Answer

It goes on increasing down the group due to decrease in ionisation energy and increase in metallic character.
Reducing power of 'Li' is highest, Na is lowest, then it goes on increasing:

Li > Cs > Rb > K > Na.

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Question 312 Marks

Eka-aluminium and eka-silicon were the names given by Mandeleev for the then unknown elements gallium and germanium respectively. A recently discovered element was first named as eka-mercury. What is its atomic number? Write its group number, electronic configuration, IUPAC and official names.

Answer

The element which comes after mercury in the periodic table is called eka-mercury.
Its various parameters are Z = 80 + 32 = 112
IUPAC name: Uub
Official name: Cn (copernicium)
Electronic configuration = $[R n] 5 f^{14} ~6 d^{10} ~7 s^2$

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Question 322 Marks

Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:

Silicon and oxygen.
Aluminium and bromine.
Calcium and iodine.
Element with atomic number 114 and fluorine.
Element with atomic number 120 and oxygen.

Answer

i. $\mathrm{SiO}_2$
ii. $\mathrm{AlBr}_3$
iii. $\mathrm{Cal}_2$
iv. $\mathrm{UuqF}_4$
v. XO where ' X ' is element with atomic number 120.

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Question 332 Marks

List the factors that can influence the ionic radius of an element.

Answer

Ionic radius is affected by following factors:

Effective nuclear charge: As the effective nuclear charge increases, ionic radius decreases.
Shielding effect: As the shielding effect increases, ionic radius increases.

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Question 342 Marks

Among the elements Li, K, Ca, S and Kr, which one is expected to have the lowest first ionisation enthalpy and which one has the highest first ionisation enthalpy?

Answer

K has lowest first ionisation enthalpy whereas Kr has highest first ionisation enthalpy.
Potassium has largest size among Li, K, Ca, S therefore, it has less force of attraction between nucleus and valence electrons, less energy is needed to remove electron.
Kr has stable outer electronic configuration, $4 \mathrm{s}^2 ~4 \mathrm{p}^6$ (octet is cmplete) therefore, it has highest ionisation enthalpy.

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Question 352 Marks

Atomic number (Z) of an element is 108. Write its electronic configuration and name the group and block to which it belongs.

Answer

$1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2 4 p^6 4 f^{14} 5 s^2 5 p^6 5 d^{10} 5 f^{14} 6 s^2 6 p^6 6 d^6 7 s^2$
It belongs to group 8 of periodic table. It belongs to d-block.

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Question 362 Marks

How many elements can be accommodated in the present set up of the long form of the periodic table? Explain.

Answer

In the present set up of the long form of the periodic table, we have eighteen groups, seven periods (i.e. principal quantum number, $n=7$ ) and four blocks ( $s, p, d$ and f-block elements).
Therefore, the maximum number of elements which can be accommodated in the present set up of the long form of the periodic table in accordance with Aufbau principle is,
$1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 4 s^2, 3 d^{10}, 4 p^6, 5 s^2, 4 d^{10}, 5 p^6, 6 s^2, 4 f^{14}, 5 d^{10}, 6 p^6, 7 s^2, 5 f^{14}, 6 d^{10}, 7 p^6=118$

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Question 372 Marks

Arrange the elements with the following electronic configurations in order of increasing electron gain enthalpy.
i. $1 s^2 2 s^2 2 p^5$
ii. $1 s^2 2 s^2 2 p^4$
iii. $1 s^2 2 s^2 2 p^3$
iv. $1 s^2 2 s^2 2 p^6 3 s^2 3 p^4$

Answer

Out of $F\left(1 s^2 2 s^2 2 p^5\right), O\left(1 s^2 2 s^2 2 p^4\right), N\left(1 s^2 2 s^2 2 p^3\right)$ and $S\left(1 s^2 2 s^2 2 p^6 3 s^2 3 p^4\right)$, only $N$ has positive electron gain enthalpy because of its stable exactly half-filled electronic configuration while all others have negative electron gain enthalpies. Since, F needs only one more electron to acquire the nearest inert gas configuration, therefore, it has the most negative electron gain enthalpy. Out of O and $\mathrm{S}, \mathrm{O}$ has less negative electron gain enthalpy than S because of electron-electron repulsions present in its small compact $2 p$-orbital. Thus, overall order of increasing electron gain enthalpy is: i $<$ (iv) $<$ (ii) $<$ (iii).

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Question 382 Marks

The electron gain enthalpy of bromine is 3.36 eV . How much energy in kcal is released when 8 g of bromine is completely converted to $\mathrm{Br}^{-}$ions in the gaseous state? $\left(1 \mathrm{eV}=23.06 \mathrm{kcal} \mathrm{~mol}^{-1}\right)$.

Answer

Number of moles of $\text{Br}=\frac{\text{Given weight}}{\text{Molecular weight}}=\frac{8}{80}=0.1\text{mol}$
Therefore, required energy $=0.1 \times 3.36 \times 23.06=7.748 \mathrm{kcal}$

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Question 392 Marks

Ionisation energy of sulphur is lower than chlorine, why?
Arrange in decreasing order of electronegativity F, O, N, Cl, C, H.

Answer

Sulphur is bigger in size than chlorine, therefore, less effective nuclear charge, less force of attraction between 16 protons and 16 electrons than 17 protons and 17 electrons in Cl, therefore it has lower ionisation enthalpy.
H < C < N < O < F is increasing order of electronegativity.

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Question 402 Marks

Lanthanoids and actinoids are placed in separate rows at the bottom of periodic table. Explain the reason for this arrangement.

Answer

Lanthanoids and actinoids resemble with each other but do not resemble any other group elements, therefore, they are kept at bottom of periodic table. They show horizontal similarity among each other.

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Question 412 Marks

The first $\left(\mathrm{IE}_1\right)$ and second $\left(\mathrm{IE}_2\right)$ ionisation energies $(\mathrm{kJ} / \mathrm{mol})$ of a new element designated by Roman numerals are shown below:

	$\text{IE}_1$	$\text{IE}_2$
I	2372	5251
II	520	7300
III	900	1760
IV	1680	3380

Which of these elements is likely to be (a) a reactive metal, (b) a reactive non metal, (c) a noble gas, and (d) a metal that forms a binary halide of the formula, $\mathrm{AX}_2$?

Answer

	$\text{IE}_1$	$\text{IE}_2$
(a)	520	7300
(b)	1680	3380
(c)	2372	5251
(d)	900	1760

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Question 422 Marks

What is formula of binary compound formed between combination of:

1st element of 1st group and iodine.
2nd element of 2nd group with 1st element of 17 group.

Answer

HI is formula.
$\mathrm{MgF}_2$ is formula.

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Question 432 Marks

Element 'Al' belongs to group 13 forms ionic compounds. Write,

Formula of its oxide.
Arrange the following in decreasing order of electropositive character Li, Na, K, Cs Give reason.

Answer

a. $\mathrm{Al}_2 \mathrm{O}_3$ is formula of its oxide.
b. $\mathrm{Cs}>\mathrm{K}>\mathrm{Na}>\mathrm{Li}$ because tendency to lose electron decreases due to decrease in atomic size and increase in effective nuclear charge.

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Question 442 Marks

${ }_{31} \mathrm{Ga}$ has slightly higher ionisation enthalpy than ${ }_{13} \mathrm{Al}$ although it occupies lower position in the group?

Answer

Electronic configuration of ${ }_{13} \mathrm{Al}=1 s^2, 2 s^2, 2 p^6, 3^{52}, 3 p^1$ Electronic configuration of $31 \mathrm{Ga}=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}$, $4 s^2, 4 p^1$
In $\mathrm{Ga}, 10$ electrons present in 3d-subshell do not shield the outer electrons from the nucleus effectively. As a result, effective nuclear charge in Ga increases. That's why ionisation enthalpy of Ga is slightly more than that of ${ }_{13} \mathrm{Al}$.

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Question 452 Marks

Justify the given statement with suitable examples- “The Properties of the elements are a periodic function of their atomic numbers”.

Answer

The similarities in the properties arise due to the same distribution of electrons in the outermost orbitals or electronic configuration which depends upon the atomic number. The elements present in a group or period exhibit similar chemical properties which depend upon the atomic number.

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Question 462 Marks

How do the electronic configurations of the elements with Z = 107 to 109 differ from one another?
Rn (Z = 86) is the last noble gas discovered. Predict what will be the atomic number of the next noble gas to be discovered. Write its symbol.

Answer

Element with Z = 107 has five, Z = 108 has six electrons while Z = 109 has seven 6d-electrons. Thus, these elements differ in the number of electrons in the 6d-subshell.
118, Uuo.

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Question 472 Marks

Why does the first ionisation energy increase as we go from left to right along a given period of periodic table? Which group has highest ionisation enthalpy?

Answer

It is due to decrease in atomic size and increase in effective nuclear charge. Group 18 because of stable electronic configuration. Group 18 because of stable electronic.

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Question 482 Marks

Arrange the following ions in the order of increasing size:
$\mathrm{Be}^{2+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-}, \mathrm{Na}^{+}, \mathrm{Mg}^{2+}, \mathrm{Br}^{-}$. Also, give reason.

Answer

$\mathrm{Be}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}<\mathrm{Cl}^{-}<\mathrm{S}^{2-}<\mathrm{Br}^{-}$
$\mathrm{Be}^{2+}$ has one shell only, $\mathrm{Mg}^{2+}$ is smaller than $\mathrm{Na}^{+}$due to more effective nuclear charge, $\mathrm{Cl}^{-}$is bigger due to less effective nuclear charge.
$\mathrm{S}^{2-}$ is bigger than $\mathrm{Cl}^{-}$due to less effective nuclear charge.
$\mathrm{Br}^{-}$is largest due to five shells (highest among these).

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Question 492 Marks

Give reason for the following statements.

Halogens act as good oxidising agents.
Electron gain enthalpy of noble gases is almost zero.

Answer

Due to highly negative electron gain enthalpy they act as good oxidising agents as they can gain electrons easily.
Electronic configuration of noble gases is such that all subshells are completely filled. Hence, their electron gain enthalpy is almost zero.

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Question 502 Marks

Predict the position of the element in the periodic table satisfying the electronic configuration (n -1) $d^1 n s^2$ for $n=4$. How many valence electrons does it here?

Answer

$3 d^1 4 s^2$ belongs to group 3 of periodic table as it has $1+2=3$ electrons in $s$ and $d$-orbitals. It has 4 valence electrons.

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Question 512 Marks

What would be IUPAC names and symbols for elements with atomic numbers 122, 127, 135, 149 and 150?

Answer

The roots 2, 7, 5, 9 and 0 are referred as bi, sept, pent, enn and nil respectively. Therefore, their names and symbol are:

Z (Atomic number)	Name	Symbol
122	Unbibium	Ubb
127	Unbiserptium	Ubs
135	Untripentium	Utp
149	Unquadennium	Uqe
150	Unpentnilium	Upn

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Question 522 Marks

How does the metallic and non metallic character vary on moving from left to right in a period?

Answer

As we move from left to right in a period, the number of valence electrons increases by one at each succeeding element but number of shells remains same. Due to this, effective nuclear charge increases. More is the effective nuclear charge, more is the attraction between the nucleus and electron.
Hence, the tendency of the element to lose electrons decreases. This results in decrease in metallic character. Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non- metallic character increases on moving from left to right in a period.

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Question 532 Marks

Name the most metallic element in second period and most non-metallic element.
Name the element with (a) largest atomic radius, (b) smallest atomic radius in third period.
Name the element having general electronic configuration $n s^2 n p^4$ in fourth period.

Answer

Most metallic element is Li and most non-metallic element is F.

Na has largest atomic radius.
Cl has smallest atomic radius.

$\operatorname{Se}(34): 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^{10} 4 p^4$ is element in fourth period with general electronic configuration $n s^2 n p^4$.

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Question 542 Marks

Arrange the following elements in increasing order of metallic character: B, Al, Mg and K.

Answer

B < Al < Mg < K are in increasing order of metallic character because atomic size increases effecive nuclear charge decreases hence, tendency to lose electrons increases.

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Question 552 Marks

What is the nature of oxides formed by most of p-block elements? Give examples.

Answer

They form mostly acidic oxides. Some of them form amphoteric and neutral oxides also. e.g. $\mathrm{CO}_2$ is acidic, $\mathrm{H}_2 \mathrm{O}$ is and CO is neutral.

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Question 562 Marks

Arrange the following elements in the increasing order of non-metallic character. Give reason.
B, C, O, N, F

Answer

B < C < N < O < F
Its is because atomic size decreases from 'B' to 'F', effective nuclear charge increases, tendency to gain electron increases, that is why non-metallic character increases.

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Question 572 Marks

Assign the position of elements having outer electronic configuration:
i. $n s^2 n p^4$ for $n=3$
ii. $(n-1) d^2 n s^2$ for $n=4$

Answer

i. $n s^2 n p^4$ for $n=3$

Since $n=3$; the element belongs to the third period. Futher, the element belongs to the $p$-block and group number 16.
ii. $(n-1) d^2 n s^2$ for $n=4$, the element belongs to period number 4 and group number 4 .

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Question 582 Marks

The reactivity of halogens decreases down the group but of alkali metal increases down the group, why?

Answer

In halogens, tendency to gain electron decreases due to increase in atomic size and less nuclear force of attraction for incoming electron. In alkali metals, tendency to lose electrons increases due to increase in atomic size and decrease in effective nuclear charge, less energy is needed to remove electron.

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Question 592 Marks

Among the elements of the third period Na to Ar, pick out the element:

With highest first ionisation enthalpy.
With largest atomic radius.
Which is most reactive non-metal.
Which is most reactive metal.

Answer

Ar has highest first ionisation enthalpy, because it is noble gas.
Na has largest atomic radius (covalent radius), it has less nuclear charge.
Cl is most reactive non-metal in third period, it can gain electron easily.
Na is most reactive metal in third period, it can lose electron easily.

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Question 602 Marks

Give the name and atomic number of the inert gas tom in which total number of d-electrons is equal to difference in number of total 'p' and s-electrons.

Answer

$\operatorname{Kr}(36) 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^{10} 4 p^6$
Number of d-electrons $=10$
Number of p-electrons $=18$
Number of s-electrons $=8$
Difference in 'p' and 's' electrons $=18-8=10$

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Question 612 Marks

Define:

Metallic radius.
Van der Waals' radius.

Answer

Metallic radius is half the distance between centres of nuclei of two atoms of metal held together by metallic bond.
Van der Waals' radius is half of the distance between centres of nuclei of two atoms held by weak van der Waals' forces of attraction.

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Question 622 Marks

How does ionisation energy vary (i) down the group, (ii) along the period from left to right?

Answer

Ionisation enthalpy decreases down in the group because atomic size increases and effective nuclear charge decreases, less energy is needed to remove electrons.
Ionisation enthalpy increases along the period from left to right due to increase in effective nuclear charge.

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Question 632 Marks

Out of O and S, which has higher negative electron gain enthalpy and why?

Answer

'S' has high electron gain enthalpy because in oxygen, there is more inter-electronic repulsion than sulphur, therefore, more energy is released in case of sulphur on gaining electrons.

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Question 642 Marks

Magnesium and Lithium both form nitride, why? Write the equation for formation of their nitride.

Answer

Li and Mg resemble with each other due to diagonal relationship i.e. due to same charge/ radius ratio that is why both form nitride.
$3 \mathrm{Mg}+\mathrm{N}_2 \rightarrow \mathrm{Mg}_3 \mathrm{N}_2$ (Magnesium nitride)
$6 \mathrm{Li}+\mathrm{N}_2 \rightarrow 2 \mathrm{Li}_3 \mathrm{N}$ (Lithium nitride)

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