2 Marks Questions

Question 12 Marks

Which of the following are isoelectronic species i.e., those having the same number of electrons?
$
Na^{+}, K^{+}, Mg^{2+}, Ca^{2+}, S^{2-}, Ar
$

Answer

$Na ^{+}$and $Mg ^{2+}$ are iso-electronic species (have 10 electrons) $K ^{+}, Ca ^{2+}, S ^{2-}$ are iso- electronic species (have 18 electrons).

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Question 22 Marks

Predict the major products of the following:
1. $CH _3- C \equiv C - Ph \xrightarrow[ H _2 O ]{ Hg ^{2+}, H ^{+}}$
2. $PhCH = CH 2 \xrightarrow{ HBr }$

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Question 32 Marks

Write structural formulae of the following compounds.
1. 3, 4, 4, 5-tetramethylheptane
2. 2, 5-dimethylhexane

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Question 42 Marks

In three moles of ethane $\left( C _2 H _6\right)$, calculate the following:
1. Number of moles of carbon atoms.
2. Number of moles of hydrogen atoms.
3. Number of molecules of ethane.

Answer

1. 1 mole of $C _2 H _6$ contains 2 moles of carbon atoms
$\therefore 3$ moles of $C _2 H _6$ will C-atoms $=2 \times 3=6$ moles
2. 1 mole of $C _2 H _6$ contains 6 moles of hydrogen atoms
$\therefore 3$ moles of $C _2 H _6$ will contain H -atoms $=3 \times 6=18$ moles
3. 1 mole of $C _2 H _6$ contains Avogadro's no., i.e. $6.02 \times 10^{23}$ molecules of ethane
$\therefore 3$ moles of $C _2 H _6$ will contain ethane molecules $=3 \times 6.02 \times 10^{23}=18.06 \times 10^{23}$

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Question 52 Marks

Would you expect the first ionization enthalpies of two isotopes of the same element to be the same or different?

Answer

Isotopes are elements with the same atomic number but different mass numbers. It implies that all isotopes of the same atom have the same electronic configurations; hence the same ionization enthalpies.

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Question 62 Marks

A sample of $HI ( g )$ is placed in a flask at a pressure of 0.2 atm . At equilibrium partial pressure of $HI ( g )$ is 0.04 atm. What is $K _{ p }$ for the given equilibrium?
$
2 H I(g) \rightleftharpoons H_2(g)+I_2(g)
$

Answer

Here, $p ( HI )=0.04 atm, P \left( H _2\right)=0.08 atm p \left( I _2\right)=0.08 atm$
Therefore, $K_p=\frac{P_{H_2} \times P_{I_2}}{\left(P_{H I}\right)^2}=\frac{(0.08 atm) \times(0.08 atm)}{(0.04 atm)^2}=4.0$

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