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JEE Chemistry 2026 Question with Solution

Given below are two statements:

Statement I: The number of species among SF4,NH4+,Ni(CO)4,XeF4,[PtCl4]2−,SeF4,[Ni(CN)4]2−SF_4, NH_4^+, Ni(CO)_4, XeF_4, [PtCl_{4}]^{2-}, SeF_4, [Ni(CN)_4]^{2-} that have tetrahedral geometry is 33.

Statement II: In the set NO,BeH2,BF3,AlCl3NO, BeH_2, BF_3, AlCl_3, all the molecules have incomplete octet around central atom.

In the light of the above statements, choose the correct answer from the options given below:

  • A

    Both Statement I and Statement II are true

  • B

    Both Statement I and Statement II are false

  • C

    Statement I is false but Statement II is true

  • D

    Statement I is true but Statement II is false

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Two statements are given about geometry and octet rule.

Find: Which statement is true or false and hence the correct option.

For Statement I, check the geometry of each species:

  • SF4SF_4 : see-saw, not tetrahedral
  • NH4+NH_4^+ : tetrahedral
  • Ni(CO)4Ni(CO)_4 : tetrahedral
  • XeF4XeF_4 : square planar
  • [PtCl4]2−[PtCl_4]^{2-} : square planar
  • SeF4SeF_4 : see-saw
  • [Ni(CN)4]2−[Ni(CN)_4]^{2-} : square planar

So, the number of tetrahedral species is only 22, namely NH4+NH_4^+ and Ni(CO)4Ni(CO)_4. Therefore, Statement I is false.

For Statement II:

  • NONO is an odd-electron species and has incomplete octet around the central atom.
  • BeH2BeH_2 has incomplete octet around Be.
  • BF3BF_3 has incomplete octet around B.
  • AlCl3AlCl_3 has incomplete octet around Al.

Thus all the given species in Statement II have incomplete octet around the central atom. Therefore, Statement II is true.

Hence, Statement I is false but Statement II is true. The correct option is C.

Geometry and Octet Analysis

Given:

  1. Statement I claims that the number of tetrahedral species is 33.
  2. Statement II claims that all species in NO,BeH2,BF3,AlCl3NO, BeH_2, BF_3, AlCl_3 have incomplete octet.

Find: Verify both statements.

Statement I: Transition metal and main group geometries are:

  • SF4SF_4 : sp3dsp^3d, see-saw
  • NH4+NH_4^+ : sp3sp^3, tetrahedral
  • Ni(CO)4Ni(CO)_4 : sp3sp^3, tetrahedral
  • XeF4XeF_4 : sp3d2sp^3d^2, square planar
  • [PtCl4]2−[PtCl_4]^{2-} : dsp2dsp^2, square planar
  • SeF4SeF_4 : sp3dsp^3d, see-saw
  • [Ni(CN)4]2−[Ni(CN)_4]^{2-} : dsp2dsp^2, square planar

Only two are tetrahedral.

Number of tetrahedral species=2\text{Number of tetrahedral species} = 2

So Statement I is false.

Statement II:

  • In NONO, the total number of valence electrons is 1111, so it is an odd-electron species and does not complete the octet on the central atom.
  • In BeH2BeH_2, Be forms two bonds and has only 44 electrons around it.
  • In BF3BF_3, B forms three bonds and has only 66 electrons around it.
  • In AlCl3AlCl_3, Al forms three bonds and has only 66 electrons around it.

Therefore, Statement II is true.

Conclude: Statement I is false but Statement II is true. Therefore, the correct option is C.

Common mistakes

  • Students often count XeF4XeF_4 as tetrahedral because it has four bonded atoms. This is wrong because molecular geometry depends on both bonded pairs and lone pairs; XeF4XeF_4 is square planar due to two lone pairs.

  • A common mistake is to assume every coordination number 44 complex is tetrahedral. This is incorrect; [PtCl4]2−[PtCl_4]^{2-} and [Ni(CN)4]2−[Ni(CN)_4]^{2-} are square planar, especially for strong-field ligands or heavier transition metals.

  • Some students think NONO satisfies octet because it contains a multiple bond. This is wrong because NONO is an odd-electron species with a total of 1111 valence electrons, so the octet remains incomplete.

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