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NEET Chemistry Solved Question Paper 2018

Multiple Choice Questions

Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s² 2s² 2p³, the simplest formula for this compound is

Mg₂X₃
MgX₂
Mg₃X₂
Mg₂X

Mg₃X₂

Element (X) electronic configuration

1s² 2s² 2p³

So, valency of X will be 3.
Valency of Mg is 2.
Formula of the compound formed by Mg and X
will be Mg₃X₂

Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity?

N₂O₅
NO₂
NO
N₂O

N₂O

Nitrous oxide (N₂O) occurs naturally in the environment.

In an automobile engine, when fuel is burnt dinitrogen and dioxygen combine to yield NO and NO₂

The bond dissociation energies of X₂, Y₂ and XY are in the ratio of 1 : 0.5: 1. ΔH for the formation of XY is –200 kJ mol–1. The bond dissociation energy of X₂ will be

200 kJ mol^-1
100 kJ mol^-1
400 kJ mol^-1
800 kJ mol^-1

800 kJ mol^-1

Let B.E of x₂,y₂ and xy are x kJ mol^-1, 0.5 x kJ mol^-1 and x kJ mol^-1 respectively

$\frac{1}{2} x_{2} + \frac{1}{2} y_{2} \to x y; ∆ H = - 200 k J m o l^{- 1} ∆ H = - 200 = Σ (B . E)_{r e a c \tan t} - Σ (B . E)_{p r o d u c t} = [\frac{1}{2} x (x) + \frac{1}{2} x (0.5) x] - [1 x (x)] O n s o l v i n g w e g e t, B . E o f X_{2} = x = 800 k J m o l^{- 1}$

Which one of the following conditions will favour maximum formation of the product in the reaction,

$A_{2} (g) + B_{2} (g) \overset{}{⇋} X_{2} (g) ∆_{r} H = - X k J ?$

Low temperature and high pressure
Low temperature and low pressure
High temperature and low pressure
High temperature and high pressure

Low temperature and high pressure

$A_{2} (g) + B_{2} (g) \overset{}{⇋} X_{2} (g); ∆ H = - x k J$

On increasing pressure equilibrium shift in a direction where number of moles decreases i.e. forward direction.

On decreasing temperature, equilibrium shift n exothermic direction i.e., forward direction.

So high pressure and lower temperature favour maximum formation of product.

5.
Which one is a wrong statement?

Total orbital angular momentum of electron in 's' orbital is equal to zero

An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers

The value of m for d_z2 is zero

The electronic configuration of N atom is

The electronic configuration of N atom is

According to Hund's Rule of maximum multiplicity, the correct electronic configuration of N-atom is

Which of the following molecules represents the order of hybridisation sp², sp², sp, sp from left to right atoms?

HC≡C-C≡CH
CH₂=CH-C≡CH
CH₃-CH=CH-CH₃
CH₂=CH-CH=CH₂

CH₂=CH-C≡CH

$\overset{s p^{2}}{C H_{2}} = \overset{s p^{2}}{C H} - \overset{s p}{C} \equiv \overset{s p}{C} H$

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :

$B r O_{4}^{-} \overset{1.82}{\to} B r O_{3}^{-} \overset{1.5 V}{\to} H B r O \overset{1.595}{\to} B r_{2} \overset{1.0652 V}{\to} B r^{-}$

Then the species undergoing disproportionation is

$B r O_{3}^{-}$
$B r O_{4}^{-}$
HBrO
Br₂

HBrO

Calculate $E_{c e l l}^{°}$ corresponding to each compound undergoing disproportionation reaction. The reaction for which $E_{c e l l}^{°}$ comes out +ve is spontaneous.

$H B r O \overset{}{\to} B r_{2} E^{o} = 1.595 V, S R P (c a t h o d e) H B r O \overset{}{\to} B r O_{3}^{-} E^{o} = - 1.5 V, S O P (a n o d e) 2 H B r O \overset{}{\to} B r_{2} + B r O_{3}^{-} E_{c e l l}^{o} = S R P (c a t h o d e) - S R P (a n o d e) = 1.595 - 1.5 = 0.095 V E_{c e l l}^{0} > 0 \Rightarrow ∆ G^{o} < 0 [s p o n \tan e o u s]$

Which of the following oxides is most acidic in nature?

BeO

In metals moving down the group metallic character increases, so basic nature increases hence most acidic will be BeO.

The correct order of atomic radii in group 13 elements is

B < Al < In < Ga < Tl
B < Al < Ga < In < Tl
B < Ga < Al < In < Tl
B < Ga < Al < Tl < In

B < Ga < Al < In < Tl

Atomic and ionic radii. Atomic and ionic radii of group 13 elements increase from top to bottom in the group. This is due to increase in the number of energy shells in each succeeding element. However, the atomic radius of gallium (Ga) is less than that of aluminium (Al).

It is due to the poor shielding of the valence electrons of Ga by the inner 3d-electrons. As a result, the effective nuclear charge of Ga is somewhat greater in magnitude than that of Al. Thus, the electrons in gallium experience the greater force of attraction by the nucleus than in aluminium. Hence the atomic size of Ga(135 pm) is slightly less than that of Al(143 pm).