Which of the following statements is/are correct?

Question

Which of the following statements is/are correct? (A) CH 3+ shows s p2- hybridization whereas CH 3- shows s p3- hybridization (B) NH4+ has a regular tetrahedral geometry (C) s p2- hybridized orbitals have equal s- and p - character (D) Hybridized orbitals always form σ bonds

Tardigrade Admin · Accepted Answer

Correct answer is (a) CH 3+ shows s p2- hybridization whereas CH 3- shows s p3- hybridizationCorrect answer is (b) NH4+ has a regular tetrahedral geometryCorrect answer is (d) Hybridized orbitals always form σ bonds

Q. Which of the following statements is/are correct?

$C H_{3}^{+}$ shows $s p^{2} -$ hybridization whereas $C H_{3}^{-}$ shows $s p^{3} -$ hybridization

$N H_{4}^{+}$ has a regular tetrahedral geometry

$s p^{2} -$ hybridized orbitals have equal s- and p - character

Hybridized orbitals always form $σ$ bonds

Correct answer is (a) $C H_{3}^{+}$ shows $s p^{2} -$ hybridization whereas $C H_{3}^{-}$ shows $s p^{3} -$ hybridization Correct answer is (b) $N H_{4}^{+}$ has a regular tetrahedral geometry Correct answer is (d) Hybridized orbitals always form $σ$ bonds

Q. Which of the following statements is/are correct?

CH3+​ shows sp2− hybridization whereas CH3−​ shows sp3− hybridization

NH4+​ has a regular tetrahedral geometry

sp2− hybridized orbitals have equal s- and p - character

Hybridized orbitals always form σ bonds

Correct answer is (a) CH3+​ shows sp2− hybridization whereas CH3−​ shows sp3− hybridization Correct answer is (b) NH4+​ has a regular tetrahedral geometry Correct answer is (d) Hybridized orbitals always form σ bonds

$C H_{3}^{+}$ shows $s p^{2} -$ hybridization whereas $C H_{3}^{-}$ shows $s p^{3} -$ hybridization

$N H_{4}^{+}$ has a regular tetrahedral geometry

$s p^{2} -$ hybridized orbitals have equal s- and p - character

Hybridized orbitals always form $σ$ bonds

Correct answer is (a) $C H_{3}^{+}$ shows $s p^{2} -$ hybridization whereas $C H_{3}^{-}$ shows $s p^{3} -$ hybridization Correct answer is (b) $N H_{4}^{+}$ has a regular tetrahedral geometry Correct answer is (d) Hybridized orbitals always form $σ$ bonds