Question 12 Marks
Which atom in $NH _4^{+}$will have formal charge +1 ?
Answer
View full question & answer→In $NH _4^{+}$, nitrogen atom ( $N$ ) will have formal charge of +1 .
Questions
Answer the following in one or two sentences.
Take a timed test
27 questions · self-marked practice — reveal the answer and mark yourself.
Question 22 Marks
Which electrons are involved in the formation of chemical bonds?
Answer
View full question & answer→The electrons present in the outermost shell of an atom are involved in the formation of a chemical bond.
Question 32 Marks
Why are atoms held together in chemical compounds?
Answer
View full question & answer→Atoms are held together in chemical compounds due to chemical bonds.
Question 42 Marks
Which part of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil?
Answer
View full question & answer→Nucleus of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil.
Question 52 Marks
Which particle was identified by J. J. Thomson in the cathode ray tube experiment?
Answer
View full question & answer→Electron was identified by J.J. Thomson in the cathode ray tube experiment.
Question 62 Marks
Does an atom have any internal structure or is it indivisible?
Answer
View full question & answer→Yes, an atom has internal structure. Different subatomic particles such as protons, electrons and neutrons constitute an atom. So, it is divisible.
Question 72 Marks
What is the difference between molecules of an element and those of a compound?
Answer
View full question & answer→The molecules of an element are made of atoms of same element while the molecules of a compound are made of atoms of different elements.
Question 82 Marks
What is the smallest unit of matter?
Answer
View full question & answer→The smallest unit of matter is atom.
Question 92 Marks
Give the names of quantum numbers.
Answer
View full question & answer→The four quantum numbers are: principal quantum number $( n )$, azimuthal or subsidiary quantum number ( 1 ), magnetic quantum number $\left( m _1\right)$ and electron spin quantum number $\left( m _{ g }\right)$.
Question 102 Marks
State Heisenberg uncertainty principle.
Answer
View full question & answer→Heisenberg uncertainty principle states that “It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron”.
Question 112 Marks
Name the element that shows simplest emission spectrum.
Answer
View full question & answer→The element that shows simplest emission spectrum is hydrogen.
Question 122 Marks
If an element ‘X’ has mass number 11 and it has 6 neutrons, then write its representation.
Answer
View full question & answer→The representation of the given element is ${ }_5^{11} X$.
Question 132 Marks
A student pictorially represented the electronic configuration of cobalt (Z = 27) in ground state as shown in the following figure.
i. Is this the correct representation?
ii. Identify the rules of electron filling that are violated (if any) in the above answer and give the correct representation.
i. Is this the correct representation?
ii. Identify the rules of electron filling that are violated (if any) in the above answer and give the correct representation.
Question 142 Marks
Write condensed orbital notation of electronic configuration of the following elements:
$i.$ Fluorine $(Z = 9)$ $ii.$ Scandium $(Z = 21)$ $iii.$ Cobalt $(Z = 27)$ $iv.$ Zinc $(Z = 30)$
$i.$ Fluorine $(Z = 9)$ $ii.$ Scandium $(Z = 21)$ $iii.$ Cobalt $(Z = 27)$ $iv.$ Zinc $(Z = 30)$
Answer
View full question & answer→| No. | Element | Condensed orbital notation |
| $i.$ | Fluorine $(Z = 9)$ | $[He] 2s^2 2p^5$ |
| $ii.$ | Scandium $(Z = 21)$ | $[Ar] 4s^2 3d^1$ |
| $iii.$ | Cobalt $(Z = 27)$ | $[Ar] 4s^2 3d^7$ |
| $iv.$ | Zinc $(Z = 30)$ | $[Ar] 4s^2 3d^{10}$ |
Question 152 Marks
What is meant by the term ‘boundary surface diagram’?
Answer
View full question & answer→A boundary surface is drawn in space for an orbital such that the value of probability density $\left(\psi^2\right)$ is constant and encloses a region where the probability of finding electron is typically more than $90 \%$. Such a boundary surface diagram is a good representation of shape of an orbital.
e.g. Boundary surface diagram of $\mathrm{Is}$ and $2 \mathrm{~s}$ orbitals are spherical in shape.
e.g. Boundary surface diagram of $\mathrm{Is}$ and $2 \mathrm{~s}$ orbitals are spherical in shape.
Question 162 Marks
Complete the following flow chart:
Question 172 Marks
Calculate the energy of a photon of radiation having wavelength $300 \ nm. [h = 6.63 \times 10^{-34} J s]$
Answer
View full question & answer→Given: Wavelength $(\lambda)=300 \mathrm{~nm}$
To find: Energy of a photon $(E)$
Formulae: $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$
Calculation: From formula,$E=\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{300 \times 10^{-9} \mathrm{~m}}=6.63 \times 10^{-19} \mathrm{~J}$
Energy of a photon is $6.63 \times 10^{-19} \mathrm{~J}$.
To find: Energy of a photon $(E)$
Formulae: $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$
Calculation: From formula,$E=\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s} \times 3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}{300 \times 10^{-9} \mathrm{~m}}=6.63 \times 10^{-19} \mathrm{~J}$
Energy of a photon is $6.63 \times 10^{-19} \mathrm{~J}$.
Question 182 Marks
What is an emission spectrum?
Answer
View full question & answer→- When a substance is irradiated with electromagnetic radiation, it absorbs energy. Atoms, molecules or ions, which have absorbed radiation are said to be ‘excited’. Heating can also result in an excited state.
- The excited species emits the absorbed energy in the form of radiation. This process is called emission of radiation and the recorded spectrum of this emitted radiation is called ‘emission spectrum’.
Question 192 Marks
Certain results were obtained when scientists studied the interactions of radiation with matter. What were the two results, utilized by Neils Bohr to overcome the drawbacks of Rutherford model?
Answer
View full question & answer→The two results utilized by Neils Bohr to overcome the drawbacks of Rutherford model were:
- Wave particle duality of electromagnetic radiation
- Line emission spectra of hydrogen
Question 202 Marks
Find out the average atomic mass of lithium (Li) from the following data:
| Isotope | Atomic mass (u) | Abundance |
| ${ }^6 \mathrm{Li}$ | 6.015 | 7.59% |
| ${ }^7 \mathrm{Li}$ | 7.016 | 92.41% |
Answer
View full question & answer→Given: Three isotopes of lithium along with respective atomic mass and % abundance.
To find: Average atomic mass of lithium
Calculation:
\begin{aligned}
\text { Average atomic mass of lithium } & =\frac{\left(\text { Atomic mass of }{ }^6 \mathrm{Li} \times \% \text { Abundance }\right)+\left(\text { Atomic mass of }{ }^7 \mathrm{Li} \times \% \text { Abundance }\right)}{100} \\
& =\frac{(6.015 \mathrm{u})(7.59)+(7.016 \mathrm{u})(92.41)}{100} \\
& =\mathbf{6 . 9 4 0} \mathbf{u}
\end{aligned}
Ans: Average atomic mass of lithium is 6.940 u.
To find: Average atomic mass of lithium
Calculation:
\begin{aligned}
\text { Average atomic mass of lithium } & =\frac{\left(\text { Atomic mass of }{ }^6 \mathrm{Li} \times \% \text { Abundance }\right)+\left(\text { Atomic mass of }{ }^7 \mathrm{Li} \times \% \text { Abundance }\right)}{100} \\
& =\frac{(6.015 \mathrm{u})(7.59)+(7.016 \mathrm{u})(92.41)}{100} \\
& =\mathbf{6 . 9 4 0} \mathbf{u}
\end{aligned}
Ans: Average atomic mass of lithium is 6.940 u.
Question 212 Marks
Complete the following table:
| Isobars | Atomic No.(Z) | Mass No.(A) | Number of Protons(Z) | No. of Neutrons(N) |
| ${ }_6^{14} \mathrm{C}$ | 6 | 14 | 6 | |
| ${ }_7^{14} \mathrm{C}$ | 7 | 14 | 7 |
Answer
View full question & answer→| Isobars | Atomic No.(Z) | Mass No.(A) | Number of Protons(Z) | No. of Neutrons(N) |
| ${ }_6^{14} \mathrm{C}$ | 6 | 14 | 6 | 8 |
| ${ }_7^{14} \mathrm{C}$ | 7 | 14 | 7 | 7 |
Question 222 Marks
Complete the information about the isotopes of carbon in the following table:
| Symbol | Atomic Number(Z) | Atomic Mass Number(A) | Neutron Number, N =A-Z | % Abundance | Stability |
| ${ }_6^{12} \mathrm{C}$ or $\mathrm{C}-12$ | 12 | 98.9% | |||
| 6 | 7 | 1.1% | Stable | ||
| ${ }_6^{14} \mathrm{C}$ or $\mathrm{C}-14$ | 6 | 14 | <0.00017% | Radioactive |
Answer
View full question & answer→| Symbol | Atomic Number(Z) | Atomic Mass Number(A) | Neutron Number, N =A-Z | % Abundance | Stability |
| ${ }_6^{12} \mathrm{C}$ or $\mathrm{C}-12$ | 6 | 12 | 6 | 98.9% | Stable |
| ${ }_6^{13} \mathrm{C}$ or $\mathrm{C}-13$ | 6 | 13 | 7 | 1.1% | Stable |
| ${ }_6^{14} \mathrm{C}$ or $\mathrm{C}-14$ | 6 | 14 | 8 | <0.00017% | Radioactive |
Question 232 Marks
Define isotopes.
Answer
View full question & answer→Isotopes are defined as the atoms of an element having the same number ofprotons but different number of neutrons in their nuclei.
e.g. ${ }_6^{12} \mathrm{C},{ }_6^{13} \mathrm{C}$ and ${ }_6^{14} \mathrm{Br}$ are isotopes.
e.g. ${ }_6^{12} \mathrm{C},{ }_6^{13} \mathrm{C}$ and ${ }_6^{14} \mathrm{Br}$ are isotopes.
Question 242 Marks
Find out the number of protons, electrons and neutrons in the nuclide ${ }_{18}^{40} \mathrm{Ar}$.
Answer
View full question & answer→For the given nuclide,
Atomic number, Z = 18, Mass number, A = 40
Number of protons = Number of electrons = Z = 18
Number of neutrons (N) = A – Z = 40 – 18 = 22
Ans: Number of protons = 18, Number of electrons = 18, Number of neutrons = 22
Atomic number, Z = 18, Mass number, A = 40
Number of protons = Number of electrons = Z = 18
Number of neutrons (N) = A – Z = 40 – 18 = 22
Ans: Number of protons = 18, Number of electrons = 18, Number of neutrons = 22
Question 252 Marks
Give reason: The approximate atomic mass in Daltons is numerically equal to the number of nucleons in the atom.
Answer
View full question & answer→- The electrons possess negligible mass. They do not contribute much to the mass of an atom.
- Therefore, the entire mass of an atom is supposed to be present in the nucleus which consists of protons and neutrons, which are collectively called as nucleons.
Question 262 Marks
Draw labelled diagram of Rutherford’s α-particle scattering experiment.
Question 272 Marks
Complete the information about the properties of subatomic particles in the following table:
| Name | Symbol | Absolute charge/C | Relative charge | Symbol for Charge | Mass/KG | Mass/u | Approximate mass/u |
| Electron | $\mathrm{e}^{-}$ | $-1.6022 \times 10^{-19}$ | $-\mathrm{e}$ | $9.10938 \times 10^{-31}$ | 0.00054 | ||
| Proton | p | +1 | $1.6726 \times 10^{-27}$ | 1.00727 | |||
| Neutron | n | 0 | $1.67493 \times 10^{-27}$ | 1.00867 | 1 |
Answer
View full question & answer→| Name | Symbol | Absolute charge/C | Relative charge | Symbol for Charge | Mass/KG | Mass/u | Approximate mass/u |
| Electron | $\mathrm{e}^{-}$ | $-1.6022 \times 10^{-19}$ | -1 | $-\mathrm{e}$ | $9.10938 \times 10^{-31}$ | 0.00054 | 0 |
| Proton | p | $+1.6022 \times 10^{-19}$ | +1 | $+\mathrm{e}$ | $1.6726 \times 10^{-27}$ | 1.00727 | 1 |
| Neutron | n | 0 | 0 | $1.67493 \times 10^{-27}$ | 1.00867 | 1 |