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Chemistry NEET Questions — Free MCQ Practice

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The transition in He⁺ ion in balmer series that would have the same wave number as the first Lyman line in the hydrogen spectrum is:

A.

2 $\to$ 1

B.

5 $\to$ 3

C.

4 $\to$ 2

D.

6 $\to$ 4

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Explanation

$For 1^{st} Lymen series of H = \frac{1}{λ} = {RZ}^{2} [\frac{1}{{(1)}^{2}} - \frac{1}{{(2)}^{2}}] For transition of {He}^{+} in Balmer series for {He}^{+} = {RZ}^{2} \frac{1}{{(2)}^{2}} - \frac{1}{{(n_{2})}^{2}}$

$For H \frac{1}{λ_{1}} = R . {(1)}^{2} . [\frac{1}{{(1)}^{2}} - \frac{1}{{(2)}^{2}}] = \frac{3 R}{4}$

$For {He}^{+} \frac{1}{λ_{2}} = R . {(2)}^{2} . \frac{1}{{(2)}^{2}} - \frac{1}{{n_{2}}^{2}} \frac{1}{λ_{2}} = 4 R \times [\frac{1}{{(2)}^{2}} - \frac{1}{{n_{2}}^{2}}]$

$Since λ_{1} = λ_{2} \frac{3 R}{4} = 4 R - [\frac{1}{4} - \frac{1}{{n_{2}}^{2}}] \frac{1}{{n_{2}}^{2}} = \frac{1}{4} - \frac{1}{16} \frac{1}{{n_{2}}^{2}} = \frac{1}{16} n_{2} = 4$

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If the speed of electron in Bohr's first orbit of hydrogen atom be x, then speed of the electron in 3rd orbit is:

A.

x/9

B.

x/3

C.

3x

D.

9x

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Explanation

$x = u_{1} = \frac{1}{1} \Rightarrow u_{1} = x F o r t h i r d o r b i t - u_{3 r d} = u_{1} \times \frac{2}{n} = x \times \frac{1}{3} = \frac{x}{3}$

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In Bohr's model of the hydrogen atom the ratio between the period of revolution of an electron in the orbit n=1 to the period of revolution of the electron in the orbit n=2 is

A.

1:2

B.

2:1

C.

1:4

D.

1:8

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Explanation

T₁/T₂ =n₁³ / n₂³ =1/8

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A cricket ball of 0.5 kg is moving with a velocity of 100 ms^-1. The wavelength associated with its motion is:

A.

1/100cm

B.

6.6 x 10^-34m

C.

1.32 x 10^-35m

D.

6.6 x 10^-28m

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Explanation

$λ = \frac{h}{m v}$ =6.6 x 10^-34/ 0.5x100 = 1.32 x 10^-35m

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How many electrons in ₁₉K have n=3; l=0?

A.

1

B.

2

C.

4

D.

3

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Explanation

₁₉K - 1S²2S²2p⁶3s²3p⁴4s¹

For 3s, n=3 & l =0

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The ratio of the difference in energy between the first and second Bohr orbit to that between the second and the third Bohr orbit is

A.

1/2

B.

1/3

C.

4/9

D.

27/5

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Explanation

$For H - atom E_{3} = - \frac{13.6}{{(3)}^{2}} = - \frac{13.6}{9} = - 1.5 eV E_{2} = - \frac{13.6}{{(2)}^{2}} = - \frac{13.6}{4} = - 3.4 eV E_{1} = - \frac{13.6}{{(1)}^{2}} = - \frac{13.6}{1} = - 13.6 eV Now, E_{2} - E_{1} = (- 3.4) - (013.6) = 13.6 - 3.4 = 10.2 eV E_{3} - E_{2} = (- 1.5) - (3.4) = 3.4 - 1.5 = 1.9 eV ∴ \frac{E_{2} - E_{1}}{E_{3} - E_{2}} = \frac{10.2}{1.9} = 5.36 \approx 5.4 = \frac{27}{5}$

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How many numbers of orbitals are possible in L-energy level?

A.

2

B.

4

C.

6

D.

1

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Explanation

L - Energy level = n=2

No. of orbitals = n² = (2)² = 4

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Which of the following is not correctly matched?

A.

Energy associated with Bohr's orbit E = $\frac{- 2.18 \times 10^{- 18} \times Z^{2}}{n^{2}}$

B.

Energy gap between two orbits $∆ E = - Z^{2} R^{} [\frac{1}{{n_{1}}^{2}} - \frac{1}{{n^{2}}_{2}}]$

C.

Kinectic energy of the ejected electron hv = hv₀ + 1/2mv²

D.

Energy of one mole of photons E = N₀hv/C

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If the velocity of a particle is reduced to 1/3^rd, then percentage increase in its de-broglie wavelength will be:-

A.

100%

B.

200%

C.

50%

D.

33.33%

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Explanation

$λ = \frac{h}{mv} λ = \frac{1}{ν} (1) ∵ λ = λ_{1} now ν = \frac{1}{3} rd λ_{2} = \frac{1}{\frac{ν}{3}} λ_{2} = \frac{3}{ν} λ_{2} = 3 \cdot λ_{1} [from (1)] % increase = \frac{λ_{2} - λ_{1}}{λ_{1}} \times 100 = \frac{3 λ_{1} - λ_{1}}{λ_{1}} \times 100 = 2 \times 100 = 200 %$

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The magnetic moment order is correctly given in

A.

Fe²⁺ < Fe³⁺ < Mn⁴⁺ < Cr³⁺

B.

Cr³⁺ = Mn⁴⁺ < Fe²⁺ < Fe³⁺

C.

Cr³⁺ = Mn⁴⁺ < Fe³⁺ < Fe²⁺

D.

Fe²⁺ < Cr³⁺ < Fe³⁺ < Mn⁴⁺

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Explanation

$μ = \sqrt{n (n + 2)}$

Fe²⁺ $\to$ 1S²2S² 2P⁶ 3S² 3P⁶ 4S² 3d⁴

n=4

Fe³⁺ $\to$ 1S²2S² 2P⁶ 3S² 3P⁶ 4S⁰ 3d⁵

n=5

Cr³⁺ $\to$ 1S²2S² 2P⁶ 3S² 3P⁶ 4S⁰ 3d³

n=3

Mn⁴⁺ $\to$ 1S²2S² 2P⁶ 3S² 3P⁶ 4S⁰ 3d³

n=3

Chemistry MCQs for NEET — Practice Questions with Answers

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