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 2011

Multiple Choice Questions

Which one of the following complexes is outer orbital complex?

[Co(NH₃)₆]³⁺
[Mn(CN)₆]⁴⁺
[Fe(CN)₆]^4-
[Ni(NH₃)₆]²⁺

[Ni(NH₃)₆]²⁺

Generally weaker field ligands form outer orbital complex. Between [Co(NH₃)₆]³⁺ and [Ni(NH₃)₆]²⁺, the later form outer orbital complex because of its d⁸ configuration. (The configuration of Co³⁺ = d⁶).

Hybridisation states of C in CH $_{3}^{+}$ and C $H_{3}^{-}$ are

sp², sp³
sp³, sp²
sp, sp²
sp², sp

sp², sp³

In CH $_{3}^{+}$ = lp + bp = 0 + 3 = 3

So, the hybridisation is sp².

In ${CH}_{3}^{-}$ = lp + bp = 1+ 3 = 4

So, the hybridisation is sp³.

Therefore, the correct answer is a.

3.
Which of the following has the maximum number of unpaired electrons?

V³⁺

Fe²⁺

Mn²⁺

Cu⁺

Mn²⁺

The outer electronic configuration of the given ions is as

10^-6 M NaOH is diluted 100 times. The pH of the diluted base is

between 7 and 8
between 5 and 6
between 6 and 7
between 10 and 11

between 7 and 8

[OH^-] in the diluted base = $\frac{10^{- 6}}{10^{2}}$ = 10^-8

Total [OH^-] = 10^-8 + [OH^-] of water

= (10^-8 + 10^-7) M

= 10^-8 [1 + 10] M

= 11 x 10^-8 M

pOH = -log (11 x 10^-8)

= - log 11 + 8 log 10

= 6.9586

pH = 14 - 6.9586 = 7.0414

For the reaction,

H₂O(l) $⇌$ H₂O(g)

at 373 K and 1 atm pressure

$∆$ H = 0
$∆$ E = 0
$∆$ H = T $∆ S$
$∆$ H = $∆$ E

$∆$ H = T $∆ S$

$∆ G = ∆ H - T ∆ S$

For the reaction,

H₂O(l) $⇌$ H₂O(g)

At equilibrium, $∆$ G = 0

Therefore, $∆ H = T ∆ S$

0.1 mol HCl is equal to

3.65 g
36.5 g
18 g
1.8 g

3.65 g

Atomic weight of Chlorine = 35.5 gm

Atomic weight of Hydrogen = 1

Molecular mass of HCl = (35.5 + 1) gm = 36.5 gm

As 1 mole is the amount of the substance which has mass equal to gram molecular mass. Therefore, 1 mol HCl is equal to 36.5 gm and 0.1 mol HCl is equal to 3.65 gm.

An sp³-hybrid orbital contains

$\frac{1}{4} s - character$
$\frac{1}{2} s - character$
$\frac{2}{3} s - character$
$\frac{3}{4} s - character$

$\frac{1}{4} s - character$

Each sp³- hybrid orbital has 25% or $\frac{1}{4}$ s- character and 75% or $\frac{3}{4}$ p- character.

The correct statement with regard to $H_{2}^{+}$ and $H_{2}^{-}$ is

both H $_{2}^{+}$ and H $_{2}^{-}$ are equally stable
both H $_{2}^{+}$ and H $_{2}^{-}$ do not exist
H $_{2}^{-}$ is more stable than H $_{2}^{+}$
H $_{2}^{+}$ is more stable than H $_{2}^{-}$

H $_{2}^{+}$ is more stable than H $_{2}^{-}$

The correct statement with regard to $H_{2}^{+}$ and $H_{2}^{-}$ is, former is more stable than later.

$H_{2}^{+} : σ 1 s^{1}, σ 1 s^{°} Bond order = \frac{1}{2} H_{2}^{-} : σ 1 s^{2}, σ 1 s^{1} Bond order = \frac{2 - 1}{2} = \frac{1}{2}$

The bond order of H $_{2}^{+}$ and H $_{2}^{-}$ are same. It is due to the presence of one electron in the antibonding molecular orbital in H $_{2}^{-}$ .

A solution made by dissolving 40 g NaOH in 1000 g of H₂O is

1 molar
1 normal
1 molal
None of these

1 molal

Number of moles NaOH (n) = $\frac{Weight in g}{molar mass}$ = $\frac{40}{40}$ = 1

As molality of a solution is defined as the number of moles of the solute dissolved in 1000 g (1 kg) of the solvent, thus the given solution of NaOH is 1 molal solution.

10.

The geometry of XeF₆ is

planar hexagon
regular octahedron
distorted octahedron
square bipyramid

distorted octahedron

The geometry of XeF₆ is distorted octahedral in which all the six positions are occupied by fluorine atoms and the lone pair of electrons of Xe atom is present at the comer of one of the triangular faces.