M(CO)x(NO)y M(NO)z
Where EAN of metal (M) in both product and reactant is same and it is 54. Then calculate value of (x + y - z). (where x, y and z are natural numbers and M belong to 6th group acc

Question

M(CO)x(NO)y $\xrightarrow{+NO}{-CO}$ M(NO)z
Where EAN of metal (M) in both product and reactant is same and it is 54. Then calculate value of (x + y - z). (where x, y and z are natural numbers and M belong to 6th group according to long form of periodic table)

Prashant Kotian · Accepted Answer

Correct answer: **ANSWER: 1**<BR>SOLUTION: METAL (M) BELONGS TO GROUP 6 AND HAS EAN = 54, SO IT IS MO (Z = 42). IN MO(CO)X(NO)Y $\RIGHTARROW$ MO(NO)Z, NO IS A 3-ELECTRON DONOR AND CO IS A 2-ELECTRON DONOR. IF Y = 2, THEN X = 3 (SINCE 42 + 3(2) + 2(3) = 54) AND Z = 4 (SINCE 42 + 4(3) = 54). HENCE, X + Y - Z = 3 + 2 - 4 = 1.. Step 1: Identify the oxidation state and configuration of the central metal in each complex ion: (a) Co^2+ is 3d^7, (b) Cr^3+ is 3d^3, (c) Fe^2+ is 3d^6, (d) Fe^3+ is 3d^5. Step 2: In [FeCl4]^2-, the central Fe^2+ (3d^6) ion is in a tetrahedral crystal field with weak-field Cl^- ligands, giving an e^3 t_2^3 configuration. Step 3: This configuration results in exactly 4 unpaired electrons. Thus, [FeCl4]^2- is the correct choice.

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