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NEET Chemistry Solved Question Paper 2014

Multiple Choice Questions

A diatomic gas at pressure P, compressed adiabatically to half of its volume, what is the final pressure?

(2)^1.4P
P/(2)^1.4
(2)^5/3P
P/(2)^5/3

(2)^1.4P

For adiabatic condition, ${PV}^{γ}$ = Constant

$P_{1}^{}$ $V_{1}^{γ}$ = $P_{2}$ $V_{2}^{γ}$ ; V₂= $\frac{1}{2}$ V₁

_{$P_{2}$ = $P_{1}$ ${[\frac{V_{1}}{V_{2}}]}^{γ}$ [For diantomic gas $γ$ = 1.4]}

_{$P_{2}$ = $P_{1}$ ${(\frac{V_{1} \times 2}{V_{1}})}^{1.4}$}

_{$P_{2}$ = $P_{1} (2)^{1.4}$} = $(2)^{1.4} P$

Which is correct regarding size of atom?

N < O
B < Ne
V > Ti
Na > K

B < Ne

The atomic radii of noble gases are by far the largest in their respective periods. This is due to the reason that noble gases have only van der Waals radii.

Which of the following represents the correct bond order?

$O_{2}^{+} < O_{2}^{-} > O_{2}^{2 -}$
$O_{2}^{-} > O_{2}^{2 -} > O_{2}^{+}$
$O_{2}^{2 -} > O_{2}^{+} > O_{2}^{-}$
$O_{2}^{+} > O_{2}^{-} > O_{2}^{2 -}$

$O_{2}^{+} > O_{2}^{-} > O_{2}^{2 -}$

Hence, the coorect Bond order is $O_{2}^{+} > O_{2}^{-} > O_{2}^{2 -}$

Ion	Total no of electron	MO configuration	Bond order
$o_{2}^{+}$	15	${KKσ2s}^{2} σ^{} 2 s^{2} σ 2 p_{x}^{2} π 2 p_{x}^{2} = π 2 p_{y}^{2} π^{} 2 p_{x}^{1} = π^{*} 2 p_{y}^{0}$	2.5
$o_{2}^{-}$	17	${KKα2s}^{2} σ^{} 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2} = π 2 p_{y}^{2} π^{} 2 p_{x}^{2} = π^{*} 2 p_{y}^{1}$	1.5
$o_{2}^{2 -}$	18	${KKσ2s}^{2} σ^{} 2 s^{2} σ 2 p_{z}^{2} π 2 p_{x}^{2} = π 2 p_{y}^{2} π^{} 2 p_{x}^{2} = π^{*} 2 p_{y}^{2}$	1.0

What is the oxidation number of Br in KBrO₄?

+ 7

Let the oxidation no. of Br be x.

In KBrO_4, + 1 + x + 4 (-2) =0,- 7 + x = 0, x=+7

5.
In O₃molecule, the formal charge on the central O-atom is

0

-1

-2

+1

Lewis gave the structure of O, molecule as

Using the relation,
Formal charge = [Total no. of valence electrons in the free atom] - [Total no. of non-bonding (lone pair) electrons]- $\frac{1}{2}$ [Total no. of bonding (shared) electrons]
The formal charge on central O -atom i.e., no. 1

=6-2- -(6)=+1

Which of the following is not sp² hybridised?

Graphite
Graphene
Fullerene
Dry ice

Dry ice

Solid CO₂ is dry ice in which carbon atomundergoes sp-hybridisation.

The equilibrium constant for the reaction,

$\frac{1}{2}$ ${H_{2}}_{(g)}$ $+$ $\frac{1}{2}$ $I_{2_{(g)}}$ $⇌$ HI_(g)is K_c

Equilibrium constant for the reaction 2HI_(g) $⇌$ H_2(g) + I_2(g) will be

1/K_c
1/(K_c)²
2/K_c
2/(K_c)²

1/(K_c)²

The given reaction is,

$\frac{1}{2} H_{2_{(g)}} + \frac{1}{2} I_{2_{(g)}} ⇋ {HI}_{(g)}$ ...(i)

Hence, $K_{C} = \frac{[HI]}{[H_{2}]^{1 / 2} [I_{2}]^{1 / 2}}$ ...(ii)

Now, reverse the equaion (i) and multiple by 2, we get

$2 HI ⇌ H_{2_{(g)}} + I_{2_{(g)}}$

$^{Hence, K_{C}^{'} = \frac{[H_{2}] [I_{2}]}{[HI]^{2}}}$

Equating equations (ii) and (iii), we get

$K_{C}^{'} = \frac{1}{(K_{C})^{2}}$

Which ofthe following statements is incorrect?

Li⁺ has minimum degree of hydration.
The oxidation state of K in KO₂ is + 1.
Na is used to make a Na/Pb alloy.
MgSO₄ is readily soluble in water.

Li⁺ has minimum degree of hydration.

The hydration enthalpies of alkali metal ions decreases with increase in ionic sizes Hence, the order is Li⁺ > Na⁺> K⁺> Rb⁺ > Cs⁺.Therefore, Li⁺ has maximum degree of hydration.