## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# NEET Chemistry Solved Question Paper 2014

#### Multiple Choice Questions

1.

Which of the following is not sp2 hybridised?

• Graphite

• Graphene

• Fullerene

• Dry ice

D.

Dry ice

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

2.

Which is correct regarding size of atom?

• N < O

• B < Ne

• V > Ti

• Na > K

B.

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.

3.

Choose the correctly paired gaseous cation and its magnetic (spin only) moment (in B.M.)

• Ti2+,3.87 B.M.

• Cr2+, 4.90 B.M.

• Co3+,3.87 B.M.

• Mn2+,4.90 B.M.

B.

Cr2+, 4.90 B.M.

Using expression, µ = $\sqrt{n\left(n+2}$ B.M. (where, n = no. of unpaired electrons)

 Ion Outer configuration n $\mu$ ${}_{22}\mathrm{Ti}^{2+}$ 3d2 2 2.84 ${}_{24}C{r}^{2+}$ 3d4 4 4.90 ${}_{27}C{o}^{3+}$ 3d6 4 4.90 ${}_{22}M{n}^{2+}$ 3d5 5 5.92

4.

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.

A.

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.

5.

Which of the following represents the correct bond order?

• ${O}_{2}^{+}<{O}_{2}^{-}>{O}_{2}^{2-}$

• ${O}_{2}^{-}>{O}_{2}^{2-}>{O}_{2}^{+}$

• ${O}_{2}^{2-}>{O}_{2}^{+}>{O}_{2}^{-}$

• ${O}_{2}^{+}>{O}_{2}^{-}>{O}_{2}^{2-}$

D.

${O}_{2}^{+}>{O}_{2}^{-}>{O}_{2}^{2-}$

Hence, the coorect Bond order is ${\text{O}}_{2}^{+}>{O}_{2}^{-}>{O}_{2}^{2-}$

 Ion Total no of electron MO configuration Bond order ${o}_{2}^{+}$ 15 2.5 ${o}_{2}^{-}$ 17 1.5 ${o}_{2}^{2-}$ 18 1.0

6.

In Omolecule, the formal charge on the central O-atom is

• 0

• -1

• -2

• +1

D.

+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

7.

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

A.

(2)1.4P

For adiabatic condition, ${\mathrm{PV}}^{\mathrm{\gamma }}$ = Constant

${\mathrm{P}}_{1}^{}$${\mathrm{V}}_{1}^{\mathrm{\gamma }}$ = ${\mathrm{P}}_{2}$${\mathrm{V}}_{2}^{\mathrm{\gamma }}$ ; V2$\frac{1}{2}$V1

${\mathrm{P}}_{2}$ = ${\mathrm{P}}_{1}$${\left[\frac{{\mathrm{V}}_{1}}{{\mathrm{V}}_{2}}\right]}^{\mathrm{\gamma }}$    [For diantomic gas $\mathrm{\gamma }$ = 1.4]

${\mathrm{P}}_{2}$ = ${\mathrm{P}}_{1}$

${\text{P}}_{2}$  =

8.

The equilibrium constant for the reaction,

$\frac{1}{2}$${{\text{H}}_{2}}_{\left(g\right)}$$+$$\frac{1}{2}$${\text{I}}_{{2}_{\left(g\right)}}$$⇌$HI(g) is Kc

Equilibrium constant for the reaction 2HI(g) $⇌$H2(g) + I2(g)  will be

• 1/Kc

• 1/(Kc)2

• 2/Kc

• 2/(Kc)2

B.

1/(Kc)2

The given reaction is,

...(i)

Hence, ${\mathrm{K}}_{\mathrm{C}}=\frac{\left[\mathrm{HI}\right]}{\left[{\mathrm{H}}_{2}{\right]}^{1/2}\left[{\mathrm{I}}_{2}{\right]}^{1/2}}$    ...(ii)

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

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

${\text{K}}_{C}^{\text{'}}=\frac{1}{\left({K}_{C}{\right)}^{2}}$

9.

Which of the following pairs represent isotones?

• ${}_{78}{}^{195}\mathrm{Pt},{}_{76}{}^{190}\mathrm{Os}$

• ${}_{47}{}^{108}\mathrm{Ag},{}_{48}{}^{112}\mathrm{Cd}$

• ${}_{72}{}^{178}\mathrm{Hf},{}_{56}{}^{137}\mathrm{Ba}$

A.

Isotones have the same number of neutrons.
As = 77 - 33 = 44 ; Se = 78 - 34 = 44

10.

What is the oxidation number of Br in KBrO?

• + 6

• + 7

• + 5

• + 8

B.

+ 7

Let the oxidation no. of Br be x.

In KBrO4, + 1 + x + 4 (-2) =0,- 7 + x = 0, x=+7