313601
Assertion : The hybridisation possessed by boron atom in \(\mathrm{BCl}_{3}\) molecule is \(\mathrm{sp}^{2}\). Reason : \(\mathrm{sp}^{2}\) hybridised molecules have trigonal planar geometry.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
313602
The percentage of s–character in the hybrid orbitals of nitrogen in \({\rm{NO}}_{\rm{2}}^{\rm{ + }}{\rm{,}}\,\,{\rm{NO}}_{\rm{3}}^{\rm{ - }}\,\,{\rm{and}}\,\,{\rm{NH}}_{\rm{4}}^{\rm{ + }}\) respectively are
1 33.3%, 25%, 50%
2 50%, 33.3%, 25%
3 25%, 50%, 33.3%
4 33.3%, 50%, 25%
Explanation:
Number of hybrid orbitals \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{V + M - C + A}}} \right)\) Where, V = Number of valence electrons in central atom M = Number of monovalent atom C = Total positive charge A = Total negative charge \({\rm{NO}}_{\rm{2}}^{\rm{ + }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{5 - 1}}} \right){\rm{ = 2}} \Rightarrow \) sp hybridisation (50% s- character) \({\rm{NO}}_{\rm{3}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{5 + 1}}} \right){\rm{ = 3}} \Rightarrow \,{\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation (33.3% s-character) \({\rm{NH}}_{\rm{4}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{5 + 4 - 1}}} \right){\rm{ = 4}} \Rightarrow {\rm{s}}{{\rm{p}}^{\rm{3}}}\) hybridisation (25% s- character)
313601
Assertion : The hybridisation possessed by boron atom in \(\mathrm{BCl}_{3}\) molecule is \(\mathrm{sp}^{2}\). Reason : \(\mathrm{sp}^{2}\) hybridised molecules have trigonal planar geometry.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
313602
The percentage of s–character in the hybrid orbitals of nitrogen in \({\rm{NO}}_{\rm{2}}^{\rm{ + }}{\rm{,}}\,\,{\rm{NO}}_{\rm{3}}^{\rm{ - }}\,\,{\rm{and}}\,\,{\rm{NH}}_{\rm{4}}^{\rm{ + }}\) respectively are
1 33.3%, 25%, 50%
2 50%, 33.3%, 25%
3 25%, 50%, 33.3%
4 33.3%, 50%, 25%
Explanation:
Number of hybrid orbitals \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{V + M - C + A}}} \right)\) Where, V = Number of valence electrons in central atom M = Number of monovalent atom C = Total positive charge A = Total negative charge \({\rm{NO}}_{\rm{2}}^{\rm{ + }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{5 - 1}}} \right){\rm{ = 2}} \Rightarrow \) sp hybridisation (50% s- character) \({\rm{NO}}_{\rm{3}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{5 + 1}}} \right){\rm{ = 3}} \Rightarrow \,{\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation (33.3% s-character) \({\rm{NH}}_{\rm{4}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left( {{\rm{5 + 4 - 1}}} \right){\rm{ = 4}} \Rightarrow {\rm{s}}{{\rm{p}}^{\rm{3}}}\) hybridisation (25% s- character)
