Subject

Chemistry

Class

NEET Class 12

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.

 Multiple Choice QuestionsMultiple Choice Questions

1.

The first ionisation enthalpy of boron is less than beryllium because:

  • Boron has high nuclear charge

  • The size of boron atom is more than beryllium atom

  • p-subshell of boron has only one electron

  • The size cit boron atom is less than beryllium atom


C.

p-subshell of boron has only one electron

Beryllium has 2s2 outer shell configuration,
while boron has a 2s2 2pconfiguration.


2.

When NaNO3 is heated in a closed vessel oxygen is liberated and NaNO3 is left behind,

At equilibrium:

  • addition of NaNO2 favours reverse reaction

  • addition of NaNO3 favours reverse reaction

  • increasing temperature favours forward reaction

  • decreasing temperature favours backward reaction


C.

increasing temperature favours forward reaction

NaNO3 (s)  NaNO2 (s) + 1 2 O2 (g)    ΔH= +veAlso Kp = (Po2)2 and thus addition of NaNO3 or NaNO2 has no effect.


3.

As the temperature is raised from 20°C to 40C, the average kinetic energy of neon atoms changes by a factor:

  • 313293

  • 313293

  • 12

  • 2


A.

313293

Avg. kinetic energy αT (kelvin)

(Factor) K·E2K·E1= T2T1= 40+27320+273= 313293

 


4.

Which of the following is the correct increasing bond order for the given diatomic species?

  • NO<O2-<C22-<He2-

  • C22-<He22+<O2-<NO

  • He2+<O2-<NO<C22-

  • O2-<NO<C22-<He2+


C.

He2+<O2-<NO<C22-

Bond order==Nb-Na2

Thus, the correct order of bond order is,

He2+<O2-<NO<C22-


5.

CaCO3   CaO + CO2  reaction in lime kiln goes to completion because:

  • CaO does not react to CO2  to give CaCO3

  • backward reaction is very low

  • CO2 formed escapes out

  • None of the above


C.

CO2 formed escapes out

In the lime kiln the CO2 formed during decomposition of calcium carbonates escape out continuously. Hence the reaction is not a reversible one and proceeds towards completion.
On the other hand if the reaction takes places in a closed container where none of the reactant or product is allowed to escape then the reaction soon reaches the equilibrium state.


6.

The dipole moment of HBr is 2.60 x 10-30 Cm and the interatomic spacing is 1.41 A .The percent ionic character of HBr is:

  • 11.5%

  • 10.1%

  • 8.2%

  • 4.1%


A.

11.5%

Theoretical value of dipole moment of a 100% ionic character

= e x d

=(1.60 x 10-19 C) (1.41 x 10-10 m)

=2.26 X 10-29 Cm

Observed value of dipole moment

=2.60 x 10-30 Cm

percent ionic character

= Observed valueTheoretical value × 100= 2.60 × 2.60 × 10-302.60 × 10-29 × 100=11.5%

 


7.

Which of the following statement(s) is/are correct regarding shielding effect?

  • The magnitude of screening effect do not depends upon the number of inner electrons

  • The magnitude of effective nuclear charge decreases in a period.

  • Greater the shielding effect, lower will be the atomic radii

  • For a given 'orbit' the shielding effect of electron belonging to different subshell decrease in the order s > p > d >f


D.

For a given 'orbit' the shielding effect of electron belonging to different subshell decrease in the order s > p > d >f

Shielding effect of all the !inner electrons is not equal. It also depends upon the subshell in which electrons are present. For a given orbit, the shielding effect of electron belonging to different subshels decreases in the order.
s>p>d>f
The screening effect of d and I-orbitals is taken as almost constant.


8.

For an electron whose x-positional uncertainity is 1x 10-10 m, the uncertainity in the x-component of the velocity in ms-1 will be of the order of:

(Data : me = 9× 10-31 Kg , h= 6.6 × 10 -34 JS )

  • 106

  • 103

  • 1012

  • 1015


A.

106

Δx · Δp  h2π            Δν n2πΔV 6.6 × 10-34 Js2×3.14×1×10-10×9×10-31ΔV 1.1 × 106 ms-1


9.

At relatively high pressure, vander Waal's equation reduces to:

  • pV=RT

  • pV=RT-aV

  • pV=RT-aV2

  • pV=RT+pb


D.

pV=RT+pb

At high pressure, the pressure correction for 1 mole of gas is negligible aV2=0

However, the volume correction cannot be neglected. Hence, van der Waal's equation reduced to,

(p+0)(V-b)=RT                [For 1 mole]

pV-pb=RT

pV=RT+pb


10.

The wavelength of a spectral line emitted by hydrogen atom in the Lyman series is 1615Rcm.
The value of n, will be

                               [R=Rydberg constant]

  • 1

  • 2

  • 3

  • 4


D.

4

For Lyman series,

1λ=R1n12-1n2215R16=R112-1n2215R16R=n22-1n22          =15 n22 =16 n22-16 n22=16, n2=4