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Chemical Bonding and Molecular Structure · Question 31

JEE Main 2025 (Online) 2nd April Evening ShiftMCQEasy📈 Topic in 23% papers+4−1
1

Which among the following molecules is (a) involved in sp3d hybridization, (b) has different bond lengths and (c) has lone pair of electrons on the central atom?

  1. XeF4
  2. XeF2
  3. PF5
  4. SF4

Answer

(4) SF4

Solution

XeF4: 4 bp + 2 lp sp3d2, square planar. Fails (a).

XeF2: 2 bp + 3 lp sp3d, linear; both Xe–F bonds are axial and equal. Fails (b).

PF5: 5 bp + 0 lp sp3d, trigonal bipyramidal; axial and equatorial bonds differ, but there is no lone pair on P. Fails (c).

SF4: 4 bp + 1 lp sp3d, see-saw shape; the lone pair is equatorial, so the axial S–F bonds are longer than the equatorial S–F bonds. Satisfies (a), (b) and (c).

Answer: (4)

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2

An element 'E' has the ionisation enthalpy value of 374 kJ mol-1. 'E' reacts with elements A, B, C and D with electron gain enthalpy values of −328, −349, −325 and −295 kJ mol-1, respectively. The correct order of the products EA, EB, EC and ED in terms of ionic character is :

  1. EB > EA > EC > ED
  2. EA > EB > EC > ED
  3. ED > EC > EA > EB
  4. ED > EC > EB > EA
JEE Main 2025 (Online) 29th January Morning Shift
3

The sum of sigma ( ) and pi ( ) bonds in Hex-1,3-dien-5-yne is .

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4

The molecules having square pyramidal geometry are

  1. BrF5 & PCl5
  2. SbF5 & PCl5
  3. BrF5 & XeOF4
  4. SbF5 & XeOF4
JEE Main 2025 (Online) 28th January Morning Shift
5

Total number of molecules/species from the following which will be paramagnetic is . O2, O2+, O2-, NO, NO2, CO, K2[NiCl4], [Co(NH3)6]Cl3, K2[Ni(CN)4]

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