Consider the following compounds with their regular geometries.
A: Square planar
B: Trigonal bipyramidal
C: Square pyramidal
D: Octahedral
E: Pentagonal bipyramidal
Calculate value o

Question

Consider the following compounds with their regular geometries.
A: Square planar
B: Trigonal bipyramidal
C: Square pyramidal
D: Octahedral
E: Pentagonal bipyramidal
Calculate value of $(x+y-z)^2$, where x is total number of axial d-orbitals having zero nodal plane, y and z are total no. of non-axial and axial d-orbitals respectively each one having two nodal planes used in hybridization of central atoms of compounds A to E.

Prashant Kotian · Accepted Answer

Correct answer: **ANSWER: 0**<BR>SOLUTION: THE HYBRIDIZATIONS ARE A ($DSP^2$, USING $D_{X^2-Y^2}$), B ($SP^3D$, USING $D_{Z^2}$), C ($SP^3D$, USING $D_{X^2-Y^2}$), D ($SP^3D^2$, USING $D_{X^2-Y^2}, D_{Z^2}$), E ($SP^3D^3$, USING $D_{X^2-Y^2}, D_{XY}$). BASED ON THE NUMBER OF NODAL PLANES: $X = D_{Z^2} = 3$, $Y = D_{XY} = 1$, $Z = D_{X^2-Y^2} = 4$. THEREFORE, $(X+Y-Z)^2 = (3+1-4)^2 = 0$.. Step 1: Identify that CN^-, CO, and NO^+ are 14-electron isoelectronic species (CN^- = 6+7+1 = 14; CO = 6+8 = 14; NO^+ = 7+8-1 = 14). Step 2: According to molecular orbital (MO) theory, any 14-electron diatomic species has a bond order of 3.0 with all electrons paired in the molecular orbitals. Step 3: Consequently, all three species are diamagnetic and possess a bond order of three, matching option (a).

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