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NEET 2021
Exam
NEET
Year
2021
Subject
Chemistry
Chapter
Coordination Compounds

NEET 2021 · Chemistry · Coordination Compounds

Match List - I with List - II. List-I List-II (a) [Fe ( CN ) _ 6 ]...

Match List - I with List - II.

List-I List-II
(a) [Fe(CN)6]3−{\left[Fe{( CN )}_{6}\right]}^{3 -} (i) 5.92BM5 . 92BM
(b) [Fe(H2O)6]3+{\left[Fe{\left({H}_{2}O\right)}_{6}\right]}^{3 +} (ii) 0BM0BM
(c) [Fe(CN)6]4−{[ Fe{( CN )}_{6} ]}^{4 -} (iii) 4.90BM4 . 90BM
(d) [Fe(H2O)6]2+{\left[Fe{\left({H}_{2}O\right)}_{6}\right]}^{2 +} (iv) 1.73BM1 . 73BM

Choose the correct answer from the options given below.

Options

  1. A(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
  2. B(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
  3. C(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
  4. D(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Correct answer

D. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Step-by-step solution

[Fe(CN)6]3−{\left[Fe ( CN{)}_{6}\right]}^{3 -} has Fe3+{Fe}^{3 +} central ion \left({d}^{5}\right)</p><p>Electronicconfiguration(SFL)</p><p></p><p>Electronic configuration (SFL)</p><p>{t}_{2g}{eg 2, 2, 1}^{0, 0}</p><p></p><p>n = 1</p><p>Magneticmoment:</p><p></p><p>Magnetic moment:</p><p>\mu = \sqrt{n\left(n + 2\right)}BM</p><p></p><p>= \sqrt{1 ( 1 + 2 )}</p><p></p><p>= \sqrt{3}</p><p></p><p>= 1 . 73BM</p><p></p><p>{\left[Fe{\left({H}_{2}O\right)}_{6}\right]}^{3 +}:hashas{d}^{5}electrons.</p><p>Electronicconfiguration(WFL)</p><p>electrons.</p><p>Electronic configuration (WFL)</p><p>{t}_{2g}{e 1, 1, 1}_{g}^{1, 1}</p><p></p><p>n = 5</p><p></p><p>\mu = \sqrt{5 ( 5 + 2 )} = \sqrt{35}</p><p></p><p>= 5 . 92B . M .

[Fe(CN)6]{\left[Fe ( CN{)}_{6}\right]}^{4 -}hashas{d}^{6}electrons.<br/>Electronicconfiguration(SFL)</p><p>electrons.<br />Electronic configuration (SFL)</p><p>{t}_{2g}^{2, 2, 1}{e}_{g}^{0, 0}</p><p></p><p>n = 0</p><p></p><p>\mu = 0B . M .

[Fe(H2O)6]2+{\left[Fe{\left({H}_{2}O\right)}_{6}\right]}^{2 +} has d6{d}^{6} electron
Electronic configuration: (WFL)

t2g2,2,1eg1,1{t}_{2g}^{2, 2, 1}{e}_{g}^{1, 1}

n=4n = 4

μ=4(4+2)=4.90B.M.</p>\mu = \sqrt{4 ( 4 + 2 )} = 4 . 90B . M .</p>

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