Chemistry MCQs for NEET — Practice Questions with Answers

(C)  v= c/${\lambda }_1$

For violet (400nm)v₁ = c / 400 x 10^-9

For red (750nm) v₂ = c/ 750x10^-9

v₁/v₂ = 750/400= 15/8

(B) Spin multiplicity = 2s+1 for minimum value no. unpaired electrons will be present so s=0.

Then Spin multiplicity = 1

(B) Ionization potential = -E₁

E₄ = E₁/16 = -50x16=E₁

Hence ionization potential = -(-800) = 800 a.u.

(A) Let x % of I(10.01) is mixed with II(11.01) and the atomic weight become 10.81. Then 

(10.01)x + (11.01)(100-x)/100 = 10.81

10.01x-11.01x+1101 = 10.81x100

or -x = 1081-1101

or -x = -20   ; x= 20

So ratio = 20% and 100-20=80%

The wave-number is defined as the reciprocal of the wavelength, and is given by ν/c, where ν is the frequency and c is the speed of light. For a frequency of 5 x 10¹⁶ Hz and c = 3 x 10⁸ m/s, the wave-number is (5 x 10¹⁶)/(3 x 10⁸) = 1.666 x 10⁸ m^-1.

(A) n =1, $\mathcal{l}$ = 2 is not valid.

(A) The central part consisting whole of the positive  charge and most of the mass caused by nucleus, it extremely small in size compared to the size of the atom.

(C) He $\underset{I.E}{\overset{I}{\to }}H{e}^{+}+e^{-}; H{e}^{+}\underset{I.E}{\overset{II}{\to }}H{e}^{2+}+e^{-}$

I(I.E)+II(I.E) = 79eV

II(I.E) = 13.6eV x Z² = 13.6x4eV = 54.4eV

I(I.E) = 79-54.4=24.6eV.

(D) Fe²⁺ has 1s²,2s²2p⁶,3s²3p⁶3d⁶

configuration with 4 unpaired electron.

(B)            Electron            Proton            Neutron        $\mathrm{\alpha }$-particle

    e            1 unit               1 unit             zero               2 units

    m          1/1837 unit        1 unit            1 unit            4 unit

   e/m           1837                1                  zero               1/2