JEE Advanced · Chemistry · 23. Coordination Compounds
Among \(\mathrm{V}(\mathrm{CO})_6, \mathrm{Cr}(\mathrm{CO})_5, \mathrm{Cu}(\mathrm{CO})_3,\)\(\operatorname{Mn}(\mathrm{CO})_5, \mathrm{Fe}(\mathrm{CO})_5,\left[\mathrm{Co}(\mathrm{CO})_3\right]^{3-},\left[\mathrm{Cr}(\mathrm{CO})_4\right]^{4-}\), and \(\operatorname{Ir}(\mathrm{CO})_3\), the total number of species isoelectronic with \(\mathrm{Ni}(\mathrm{CO})_4\) is ________.
[Given, atomic number:\(\mathrm{V}=23, \mathrm{Cr}=24, \mathrm{Mn}=25,\) \(\mathrm{Fe}=26, \mathrm{Co}=27, \mathrm{Ni}=28, \mathrm{Cu}=29,\) \(\mathrm{Ir}=77\)]
- A 3
- B 5
- C 7
- D 9
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
In case of complexes, isoelectronic species should be those having same effective atomic number (EAN)
\(\mathrm{Ni}(\mathrm{CO})_4 \Rightarrow 28+4 \times 2=36\)
(i) \(\mathrm{V}(\mathrm{CO})_6 \Rightarrow 23+2 \times 6=35\)
(ii) \(\operatorname{Cr}(\mathrm{CO})_5 \Rightarrow 24+2 \times 5=34\)
(iii) \(\mathrm{Cu}(\mathrm{CO})_3 \Rightarrow 29+2 \times 3=35\)
(iv) \(\mathrm{Mn}(\mathrm{CO})_5 \Rightarrow 25+2 \times 5=35\)
(v) \(\mathrm{Fe}(\mathrm{CO})_5 \Rightarrow 26+2 \times 5=36\)
(vi) \(\left[\mathrm{Co}(\mathrm{CO})_3\right]^{3-} \Rightarrow 27+3+2 \times 3=36\)
(vii) \(\left[\mathrm{Cr}(\mathrm{CO})_4\right]^{4-} \Rightarrow 24+4+2 \times 4=36\)
(viii) \(\left[\operatorname{Ir}(\mathrm{CO})_3\right] \Rightarrow 77+2 \times 3=83\)
\(\mathrm{Ni}(\mathrm{CO})_4 \Rightarrow 28+4 \times 2=36\)
(i) \(\mathrm{V}(\mathrm{CO})_6 \Rightarrow 23+2 \times 6=35\)
(ii) \(\operatorname{Cr}(\mathrm{CO})_5 \Rightarrow 24+2 \times 5=34\)
(iii) \(\mathrm{Cu}(\mathrm{CO})_3 \Rightarrow 29+2 \times 3=35\)
(iv) \(\mathrm{Mn}(\mathrm{CO})_5 \Rightarrow 25+2 \times 5=35\)
(v) \(\mathrm{Fe}(\mathrm{CO})_5 \Rightarrow 26+2 \times 5=36\)
(vi) \(\left[\mathrm{Co}(\mathrm{CO})_3\right]^{3-} \Rightarrow 27+3+2 \times 3=36\)
(vii) \(\left[\mathrm{Cr}(\mathrm{CO})_4\right]^{4-} \Rightarrow 24+4+2 \times 4=36\)
(viii) \(\left[\operatorname{Ir}(\mathrm{CO})_3\right] \Rightarrow 77+2 \times 3=83\)
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