313573
The species in which the N atom is in a state of sp hybridisation is:
1 \({\rm{NO}}_{\rm{3}}^{\rm{ - }}\)
2 \({\rm{N}}{{\rm{O}}_{\rm{2}}}\)
3 \({\rm{NO}}_{\rm{2}}^{\rm{ + }}\)
4 \({\rm{NO}}_{\rm{2}}^{\rm{ - }}\)
Explanation:
Hybridisation \(\left( {\rm{H}} \right){\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\) [no.of valence electron of central atom + no.of Monovalent atoms attached to it + (-ve charge if any ) - ( +ve charge if any)] \({\rm{NO}}_{\rm{2}}^{\rm{ + }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 0 - 1}}} \right]{\rm{ = 2}}\) i.e., sp hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation The lewis structure of \({\rm{N}}{{\rm{O}}_{\rm{2}}}\) shows a bent molecular geometry with trigonal planar electron pair geometry hence the hybridisation will be \({\rm{s}}{{\rm{p}}^{\rm{2}}}\).
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313574
What is hybridisation of \({\rm{'O'}}\) in \(\mathrm{H}_{2} \mathrm{O}\) ?
1 \(\mathrm{sp}\)
2 \(\mathrm{sp}^{3}\)
3 \(\mathrm{sp}^{2}\)
4 No hybridisation
Explanation:
In gaseous state, water exists as discrete \(\mathrm{H}_{2} \mathrm{O}\) molecule, in which the central \(\mathrm{O}\)-atom contains four \(\mathrm{sp}^{3}\)-hybridised orbitals, two of which are half - filled and used in overlapping with 1s orbital of \(\mathrm{H}\)-atom to give two \(\mathrm{sp}^{3}-\mathrm{s}, \mathrm{O}-\mathrm{H}, \sigma\)-bond while the remaining two exist as lone pair. The geometry of the molecule is angular or \(\mathrm{V}\)-shape.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313575
Number of hybrid orbitals present in a molecule of propene are
1 12
2 10
3 9
4 8
Explanation:
\(\mathop {{\text{C}}{{\text{H}}_3}}\limits_{\left( {{\text{s}}{{\text{p}}^3}} \right)} - \mathop {{\text{CH}}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} = \mathop {{\text{C}}{{\text{H}}_2}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} \). So, it has \(4+3+3=10\) hybrid orbitals.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313576
All the four sigma bonds in perchlorate ion are
In perchlorate ion, Central \(\mathrm{Cl}\)-atom is \(\mathrm{sp}^{3}\) hybridised. So, all the our sigma bonds are \(\mathrm{sp}^{3}-\mathrm{p}\) bonds.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313577
If a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) has the dipole moment zero, then the sigma \({\rm{(\sigma )}}\) bonding orbitals used by \({\text{M(Z}} < {\text{21}})\) is
1 \({\rm{s}}{{\rm{p}}^{\rm{2}}}\)-hybridised
2 pure p-orbitals
3 \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridised
4 sp-hybridised
Explanation:
In a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) with zero dipole moment the sigma bonding orbitals used by 'M' is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) - hybridised. e.g., \(\mathrm{BF}_{3}, \mathrm{BCl}_{3}\), etc.
313573
The species in which the N atom is in a state of sp hybridisation is:
1 \({\rm{NO}}_{\rm{3}}^{\rm{ - }}\)
2 \({\rm{N}}{{\rm{O}}_{\rm{2}}}\)
3 \({\rm{NO}}_{\rm{2}}^{\rm{ + }}\)
4 \({\rm{NO}}_{\rm{2}}^{\rm{ - }}\)
Explanation:
Hybridisation \(\left( {\rm{H}} \right){\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\) [no.of valence electron of central atom + no.of Monovalent atoms attached to it + (-ve charge if any ) - ( +ve charge if any)] \({\rm{NO}}_{\rm{2}}^{\rm{ + }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 0 - 1}}} \right]{\rm{ = 2}}\) i.e., sp hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation The lewis structure of \({\rm{N}}{{\rm{O}}_{\rm{2}}}\) shows a bent molecular geometry with trigonal planar electron pair geometry hence the hybridisation will be \({\rm{s}}{{\rm{p}}^{\rm{2}}}\).
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313574
What is hybridisation of \({\rm{'O'}}\) in \(\mathrm{H}_{2} \mathrm{O}\) ?
1 \(\mathrm{sp}\)
2 \(\mathrm{sp}^{3}\)
3 \(\mathrm{sp}^{2}\)
4 No hybridisation
Explanation:
In gaseous state, water exists as discrete \(\mathrm{H}_{2} \mathrm{O}\) molecule, in which the central \(\mathrm{O}\)-atom contains four \(\mathrm{sp}^{3}\)-hybridised orbitals, two of which are half - filled and used in overlapping with 1s orbital of \(\mathrm{H}\)-atom to give two \(\mathrm{sp}^{3}-\mathrm{s}, \mathrm{O}-\mathrm{H}, \sigma\)-bond while the remaining two exist as lone pair. The geometry of the molecule is angular or \(\mathrm{V}\)-shape.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313575
Number of hybrid orbitals present in a molecule of propene are
1 12
2 10
3 9
4 8
Explanation:
\(\mathop {{\text{C}}{{\text{H}}_3}}\limits_{\left( {{\text{s}}{{\text{p}}^3}} \right)} - \mathop {{\text{CH}}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} = \mathop {{\text{C}}{{\text{H}}_2}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} \). So, it has \(4+3+3=10\) hybrid orbitals.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313576
All the four sigma bonds in perchlorate ion are
In perchlorate ion, Central \(\mathrm{Cl}\)-atom is \(\mathrm{sp}^{3}\) hybridised. So, all the our sigma bonds are \(\mathrm{sp}^{3}-\mathrm{p}\) bonds.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313577
If a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) has the dipole moment zero, then the sigma \({\rm{(\sigma )}}\) bonding orbitals used by \({\text{M(Z}} < {\text{21}})\) is
1 \({\rm{s}}{{\rm{p}}^{\rm{2}}}\)-hybridised
2 pure p-orbitals
3 \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridised
4 sp-hybridised
Explanation:
In a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) with zero dipole moment the sigma bonding orbitals used by 'M' is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) - hybridised. e.g., \(\mathrm{BF}_{3}, \mathrm{BCl}_{3}\), etc.
313573
The species in which the N atom is in a state of sp hybridisation is:
1 \({\rm{NO}}_{\rm{3}}^{\rm{ - }}\)
2 \({\rm{N}}{{\rm{O}}_{\rm{2}}}\)
3 \({\rm{NO}}_{\rm{2}}^{\rm{ + }}\)
4 \({\rm{NO}}_{\rm{2}}^{\rm{ - }}\)
Explanation:
Hybridisation \(\left( {\rm{H}} \right){\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\) [no.of valence electron of central atom + no.of Monovalent atoms attached to it + (-ve charge if any ) - ( +ve charge if any)] \({\rm{NO}}_{\rm{2}}^{\rm{ + }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 0 - 1}}} \right]{\rm{ = 2}}\) i.e., sp hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation The lewis structure of \({\rm{N}}{{\rm{O}}_{\rm{2}}}\) shows a bent molecular geometry with trigonal planar electron pair geometry hence the hybridisation will be \({\rm{s}}{{\rm{p}}^{\rm{2}}}\).
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313574
What is hybridisation of \({\rm{'O'}}\) in \(\mathrm{H}_{2} \mathrm{O}\) ?
1 \(\mathrm{sp}\)
2 \(\mathrm{sp}^{3}\)
3 \(\mathrm{sp}^{2}\)
4 No hybridisation
Explanation:
In gaseous state, water exists as discrete \(\mathrm{H}_{2} \mathrm{O}\) molecule, in which the central \(\mathrm{O}\)-atom contains four \(\mathrm{sp}^{3}\)-hybridised orbitals, two of which are half - filled and used in overlapping with 1s orbital of \(\mathrm{H}\)-atom to give two \(\mathrm{sp}^{3}-\mathrm{s}, \mathrm{O}-\mathrm{H}, \sigma\)-bond while the remaining two exist as lone pair. The geometry of the molecule is angular or \(\mathrm{V}\)-shape.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313575
Number of hybrid orbitals present in a molecule of propene are
1 12
2 10
3 9
4 8
Explanation:
\(\mathop {{\text{C}}{{\text{H}}_3}}\limits_{\left( {{\text{s}}{{\text{p}}^3}} \right)} - \mathop {{\text{CH}}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} = \mathop {{\text{C}}{{\text{H}}_2}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} \). So, it has \(4+3+3=10\) hybrid orbitals.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313576
All the four sigma bonds in perchlorate ion are
In perchlorate ion, Central \(\mathrm{Cl}\)-atom is \(\mathrm{sp}^{3}\) hybridised. So, all the our sigma bonds are \(\mathrm{sp}^{3}-\mathrm{p}\) bonds.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313577
If a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) has the dipole moment zero, then the sigma \({\rm{(\sigma )}}\) bonding orbitals used by \({\text{M(Z}} < {\text{21}})\) is
1 \({\rm{s}}{{\rm{p}}^{\rm{2}}}\)-hybridised
2 pure p-orbitals
3 \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridised
4 sp-hybridised
Explanation:
In a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) with zero dipole moment the sigma bonding orbitals used by 'M' is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) - hybridised. e.g., \(\mathrm{BF}_{3}, \mathrm{BCl}_{3}\), etc.
313573
The species in which the N atom is in a state of sp hybridisation is:
1 \({\rm{NO}}_{\rm{3}}^{\rm{ - }}\)
2 \({\rm{N}}{{\rm{O}}_{\rm{2}}}\)
3 \({\rm{NO}}_{\rm{2}}^{\rm{ + }}\)
4 \({\rm{NO}}_{\rm{2}}^{\rm{ - }}\)
Explanation:
Hybridisation \(\left( {\rm{H}} \right){\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\) [no.of valence electron of central atom + no.of Monovalent atoms attached to it + (-ve charge if any ) - ( +ve charge if any)] \({\rm{NO}}_{\rm{2}}^{\rm{ + }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 0 - 1}}} \right]{\rm{ = 2}}\) i.e., sp hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation The lewis structure of \({\rm{N}}{{\rm{O}}_{\rm{2}}}\) shows a bent molecular geometry with trigonal planar electron pair geometry hence the hybridisation will be \({\rm{s}}{{\rm{p}}^{\rm{2}}}\).
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313574
What is hybridisation of \({\rm{'O'}}\) in \(\mathrm{H}_{2} \mathrm{O}\) ?
1 \(\mathrm{sp}\)
2 \(\mathrm{sp}^{3}\)
3 \(\mathrm{sp}^{2}\)
4 No hybridisation
Explanation:
In gaseous state, water exists as discrete \(\mathrm{H}_{2} \mathrm{O}\) molecule, in which the central \(\mathrm{O}\)-atom contains four \(\mathrm{sp}^{3}\)-hybridised orbitals, two of which are half - filled and used in overlapping with 1s orbital of \(\mathrm{H}\)-atom to give two \(\mathrm{sp}^{3}-\mathrm{s}, \mathrm{O}-\mathrm{H}, \sigma\)-bond while the remaining two exist as lone pair. The geometry of the molecule is angular or \(\mathrm{V}\)-shape.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313575
Number of hybrid orbitals present in a molecule of propene are
1 12
2 10
3 9
4 8
Explanation:
\(\mathop {{\text{C}}{{\text{H}}_3}}\limits_{\left( {{\text{s}}{{\text{p}}^3}} \right)} - \mathop {{\text{CH}}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} = \mathop {{\text{C}}{{\text{H}}_2}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} \). So, it has \(4+3+3=10\) hybrid orbitals.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313576
All the four sigma bonds in perchlorate ion are
In perchlorate ion, Central \(\mathrm{Cl}\)-atom is \(\mathrm{sp}^{3}\) hybridised. So, all the our sigma bonds are \(\mathrm{sp}^{3}-\mathrm{p}\) bonds.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313577
If a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) has the dipole moment zero, then the sigma \({\rm{(\sigma )}}\) bonding orbitals used by \({\text{M(Z}} < {\text{21}})\) is
1 \({\rm{s}}{{\rm{p}}^{\rm{2}}}\)-hybridised
2 pure p-orbitals
3 \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridised
4 sp-hybridised
Explanation:
In a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) with zero dipole moment the sigma bonding orbitals used by 'M' is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) - hybridised. e.g., \(\mathrm{BF}_{3}, \mathrm{BCl}_{3}\), etc.
313573
The species in which the N atom is in a state of sp hybridisation is:
1 \({\rm{NO}}_{\rm{3}}^{\rm{ - }}\)
2 \({\rm{N}}{{\rm{O}}_{\rm{2}}}\)
3 \({\rm{NO}}_{\rm{2}}^{\rm{ + }}\)
4 \({\rm{NO}}_{\rm{2}}^{\rm{ - }}\)
Explanation:
Hybridisation \(\left( {\rm{H}} \right){\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\) [no.of valence electron of central atom + no.of Monovalent atoms attached to it + (-ve charge if any ) - ( +ve charge if any)] \({\rm{NO}}_{\rm{2}}^{\rm{ + }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 0 - 1}}} \right]{\rm{ = 2}}\) i.e., sp hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation \({\rm{NO}}_{\rm{2}}^{\rm{ - }}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\left[ {{\rm{5 + 0 + 1 - 0}}} \right]{\rm{ = 3}}\) i.e., \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridisation The lewis structure of \({\rm{N}}{{\rm{O}}_{\rm{2}}}\) shows a bent molecular geometry with trigonal planar electron pair geometry hence the hybridisation will be \({\rm{s}}{{\rm{p}}^{\rm{2}}}\).
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313574
What is hybridisation of \({\rm{'O'}}\) in \(\mathrm{H}_{2} \mathrm{O}\) ?
1 \(\mathrm{sp}\)
2 \(\mathrm{sp}^{3}\)
3 \(\mathrm{sp}^{2}\)
4 No hybridisation
Explanation:
In gaseous state, water exists as discrete \(\mathrm{H}_{2} \mathrm{O}\) molecule, in which the central \(\mathrm{O}\)-atom contains four \(\mathrm{sp}^{3}\)-hybridised orbitals, two of which are half - filled and used in overlapping with 1s orbital of \(\mathrm{H}\)-atom to give two \(\mathrm{sp}^{3}-\mathrm{s}, \mathrm{O}-\mathrm{H}, \sigma\)-bond while the remaining two exist as lone pair. The geometry of the molecule is angular or \(\mathrm{V}\)-shape.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313575
Number of hybrid orbitals present in a molecule of propene are
1 12
2 10
3 9
4 8
Explanation:
\(\mathop {{\text{C}}{{\text{H}}_3}}\limits_{\left( {{\text{s}}{{\text{p}}^3}} \right)} - \mathop {{\text{CH}}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} = \mathop {{\text{C}}{{\text{H}}_2}}\limits_{\left( {{\text{s}}{{\text{p}}^2}} \right)} \). So, it has \(4+3+3=10\) hybrid orbitals.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313576
All the four sigma bonds in perchlorate ion are
In perchlorate ion, Central \(\mathrm{Cl}\)-atom is \(\mathrm{sp}^{3}\) hybridised. So, all the our sigma bonds are \(\mathrm{sp}^{3}-\mathrm{p}\) bonds.
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313577
If a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) has the dipole moment zero, then the sigma \({\rm{(\sigma )}}\) bonding orbitals used by \({\text{M(Z}} < {\text{21}})\) is
1 \({\rm{s}}{{\rm{p}}^{\rm{2}}}\)-hybridised
2 pure p-orbitals
3 \({\rm{s}}{{\rm{p}}^{\rm{3}}}\)-hybridised
4 sp-hybridised
Explanation:
In a molecule \({\rm{M}}{{\rm{X}}_{\rm{3}}}{\rm{ }}\) with zero dipole moment the sigma bonding orbitals used by 'M' is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) - hybridised. e.g., \(\mathrm{BF}_{3}, \mathrm{BCl}_{3}\), etc.
