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JEE Advanced Chemistry Coordination Compounds 2024 JEE Advanced 2024 (Paper 1)

JEE Advanced Chemistry Question (2024) — Solution

Question

Among V ( CO )_6, Cr ( CO )_5, Cu ( CO )_3, Mn ( CO )_5, Fe ( CO )_5, [ Co ( CO )_3 ]^ 3- , [ Cr ( CO )_4 ]^ 4- , and Ir ( CO )_3, the total number of species isoelectronic with Ni ( CO )_4 is ________. [Given, atomic number: V =23, Cr =24, Mn =25, Fe =26, Co =27, Ni =28, Cu =29, Ir =77]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

In case of complexes, isoelectronic species should be those having same effective atomic number (EAN) Ni ( CO )_4 28+4 2=36 (i) V ( CO )_6 23+2 6=35 (ii) Cr ( CO )_5 24+2 5=34 (iii) Cu ( CO )_3 29+2 3=35 (iv) Mn ( CO )_5 25+2 5=35 (v) Fe ( CO )_5 26+2 5=36 (vi) [ Co ( CO )_3 ]^ 3- 27+3+2 3=36 (vii) [ Cr ( CO )_4 ]^ 4- 24+4+2 4=36 (viii) [ Ir ( CO )_3 ] 77+2 3=83

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