285254
The formal charge on central oxygen atom in ozone is
1 0
2 +2
3 +1
4 -1
Explanation:
(c)Formal charge
\(=\) Total number of valence electrons
- Total number of non-bonding electron
\(-\frac{1}{2}\) [Total number of bonding electrons]
Formal charge on O -atom numbered 1
\(=6-4-\frac{4}{2}=0\)
Formal charge on O -atom numbered 2
\(=6-2-\frac{6}{2}=+1\)
Formal charge on O -atom numbered 3
\(=6-6-\frac{2}{2}=-1\)
Karnataka CET 2020
CHEMISTRY(KCET)
285255
The percentage of\(s\)-character in the hybrid orbitals of nitrogen in \(\mathrm{NO}_2^{+}, \mathrm{NO}_3^{-}\)and \(\mathrm{NO}_4^{+}\)respectively are
1 \(33.3 \%, 25 \%, 50 \%\)
2 \(50 \%, 33.3 \%, 25 \%\)
3 \(25 \%, 50 \%, 33.3 \%\)
4 \(33.3 \%, 50 \%, 25 \%\)
Explanation:
(b) Number of hybrid orbitals
\(=\frac{1}{2}(V+M-C+A)\)
Where, \(V=\) Number of valence electrons in central atom
\(M=\) Number of monovalent atom
\(C=\) Total positive charge
\(A=\) Total negative charge
\(\mathrm{NO}_2^{+}=\frac{1}{2}(5-1)=2 \Rightarrow s p\) hybridisaiton
(50\% s-character)
\(\mathrm{NO}_3^{-}=\frac{1}{2}(5+1)=3 \Rightarrow s p^2\) hybridisation
(33.3\% \(s\)-character)
\(\mathrm{NH}_4^{+}=\frac{1}{2}(5+4-1)=4 \Rightarrow s p^3\) hybridisation
(25\% \(s\)-character)
Karnataka CET 2020
CHEMISTRY(KCET)
285256
Bond angle in\(\mathrm{PH}_4^{+}\)is more than that of \(\mathrm{PH}_3\). This is because
3 hybridisation of P changes when \(\mathrm{PH}_3\) is converted to \(\mathrm{PH}_4^{+}\)
4 lone pair - bond pair repulsion exists in \(\mathrm{PH}_3\).
Explanation:
(d)
Both are \(s p^3\) hybridised. In \(\mathrm{PH}_4^{+}\), all the four orbitals are bonded whereas in \(\mathrm{PH}_3\) there is a lone pair of electrons on P , which is responsible for lone pairbond pair repulsion in \(\mathrm{PH}_3\) reducing the bond angle to less than \(109^{\circ} 28^{\prime}\).
285254
The formal charge on central oxygen atom in ozone is
1 0
2 +2
3 +1
4 -1
Explanation:
(c)Formal charge
\(=\) Total number of valence electrons
- Total number of non-bonding electron
\(-\frac{1}{2}\) [Total number of bonding electrons]
Formal charge on O -atom numbered 1
\(=6-4-\frac{4}{2}=0\)
Formal charge on O -atom numbered 2
\(=6-2-\frac{6}{2}=+1\)
Formal charge on O -atom numbered 3
\(=6-6-\frac{2}{2}=-1\)
Karnataka CET 2020
CHEMISTRY(KCET)
285255
The percentage of\(s\)-character in the hybrid orbitals of nitrogen in \(\mathrm{NO}_2^{+}, \mathrm{NO}_3^{-}\)and \(\mathrm{NO}_4^{+}\)respectively are
1 \(33.3 \%, 25 \%, 50 \%\)
2 \(50 \%, 33.3 \%, 25 \%\)
3 \(25 \%, 50 \%, 33.3 \%\)
4 \(33.3 \%, 50 \%, 25 \%\)
Explanation:
(b) Number of hybrid orbitals
\(=\frac{1}{2}(V+M-C+A)\)
Where, \(V=\) Number of valence electrons in central atom
\(M=\) Number of monovalent atom
\(C=\) Total positive charge
\(A=\) Total negative charge
\(\mathrm{NO}_2^{+}=\frac{1}{2}(5-1)=2 \Rightarrow s p\) hybridisaiton
(50\% s-character)
\(\mathrm{NO}_3^{-}=\frac{1}{2}(5+1)=3 \Rightarrow s p^2\) hybridisation
(33.3\% \(s\)-character)
\(\mathrm{NH}_4^{+}=\frac{1}{2}(5+4-1)=4 \Rightarrow s p^3\) hybridisation
(25\% \(s\)-character)
Karnataka CET 2020
CHEMISTRY(KCET)
285256
Bond angle in\(\mathrm{PH}_4^{+}\)is more than that of \(\mathrm{PH}_3\). This is because
3 hybridisation of P changes when \(\mathrm{PH}_3\) is converted to \(\mathrm{PH}_4^{+}\)
4 lone pair - bond pair repulsion exists in \(\mathrm{PH}_3\).
Explanation:
(d)
Both are \(s p^3\) hybridised. In \(\mathrm{PH}_4^{+}\), all the four orbitals are bonded whereas in \(\mathrm{PH}_3\) there is a lone pair of electrons on P , which is responsible for lone pairbond pair repulsion in \(\mathrm{PH}_3\) reducing the bond angle to less than \(109^{\circ} 28^{\prime}\).
285254
The formal charge on central oxygen atom in ozone is
1 0
2 +2
3 +1
4 -1
Explanation:
(c)Formal charge
\(=\) Total number of valence electrons
- Total number of non-bonding electron
\(-\frac{1}{2}\) [Total number of bonding electrons]
Formal charge on O -atom numbered 1
\(=6-4-\frac{4}{2}=0\)
Formal charge on O -atom numbered 2
\(=6-2-\frac{6}{2}=+1\)
Formal charge on O -atom numbered 3
\(=6-6-\frac{2}{2}=-1\)
Karnataka CET 2020
CHEMISTRY(KCET)
285255
The percentage of\(s\)-character in the hybrid orbitals of nitrogen in \(\mathrm{NO}_2^{+}, \mathrm{NO}_3^{-}\)and \(\mathrm{NO}_4^{+}\)respectively are
1 \(33.3 \%, 25 \%, 50 \%\)
2 \(50 \%, 33.3 \%, 25 \%\)
3 \(25 \%, 50 \%, 33.3 \%\)
4 \(33.3 \%, 50 \%, 25 \%\)
Explanation:
(b) Number of hybrid orbitals
\(=\frac{1}{2}(V+M-C+A)\)
Where, \(V=\) Number of valence electrons in central atom
\(M=\) Number of monovalent atom
\(C=\) Total positive charge
\(A=\) Total negative charge
\(\mathrm{NO}_2^{+}=\frac{1}{2}(5-1)=2 \Rightarrow s p\) hybridisaiton
(50\% s-character)
\(\mathrm{NO}_3^{-}=\frac{1}{2}(5+1)=3 \Rightarrow s p^2\) hybridisation
(33.3\% \(s\)-character)
\(\mathrm{NH}_4^{+}=\frac{1}{2}(5+4-1)=4 \Rightarrow s p^3\) hybridisation
(25\% \(s\)-character)
Karnataka CET 2020
CHEMISTRY(KCET)
285256
Bond angle in\(\mathrm{PH}_4^{+}\)is more than that of \(\mathrm{PH}_3\). This is because
3 hybridisation of P changes when \(\mathrm{PH}_3\) is converted to \(\mathrm{PH}_4^{+}\)
4 lone pair - bond pair repulsion exists in \(\mathrm{PH}_3\).
Explanation:
(d)
Both are \(s p^3\) hybridised. In \(\mathrm{PH}_4^{+}\), all the four orbitals are bonded whereas in \(\mathrm{PH}_3\) there is a lone pair of electrons on P , which is responsible for lone pairbond pair repulsion in \(\mathrm{PH}_3\) reducing the bond angle to less than \(109^{\circ} 28^{\prime}\).
285254
The formal charge on central oxygen atom in ozone is
1 0
2 +2
3 +1
4 -1
Explanation:
(c)Formal charge
\(=\) Total number of valence electrons
- Total number of non-bonding electron
\(-\frac{1}{2}\) [Total number of bonding electrons]
Formal charge on O -atom numbered 1
\(=6-4-\frac{4}{2}=0\)
Formal charge on O -atom numbered 2
\(=6-2-\frac{6}{2}=+1\)
Formal charge on O -atom numbered 3
\(=6-6-\frac{2}{2}=-1\)
Karnataka CET 2020
CHEMISTRY(KCET)
285255
The percentage of\(s\)-character in the hybrid orbitals of nitrogen in \(\mathrm{NO}_2^{+}, \mathrm{NO}_3^{-}\)and \(\mathrm{NO}_4^{+}\)respectively are
1 \(33.3 \%, 25 \%, 50 \%\)
2 \(50 \%, 33.3 \%, 25 \%\)
3 \(25 \%, 50 \%, 33.3 \%\)
4 \(33.3 \%, 50 \%, 25 \%\)
Explanation:
(b) Number of hybrid orbitals
\(=\frac{1}{2}(V+M-C+A)\)
Where, \(V=\) Number of valence electrons in central atom
\(M=\) Number of monovalent atom
\(C=\) Total positive charge
\(A=\) Total negative charge
\(\mathrm{NO}_2^{+}=\frac{1}{2}(5-1)=2 \Rightarrow s p\) hybridisaiton
(50\% s-character)
\(\mathrm{NO}_3^{-}=\frac{1}{2}(5+1)=3 \Rightarrow s p^2\) hybridisation
(33.3\% \(s\)-character)
\(\mathrm{NH}_4^{+}=\frac{1}{2}(5+4-1)=4 \Rightarrow s p^3\) hybridisation
(25\% \(s\)-character)
Karnataka CET 2020
CHEMISTRY(KCET)
285256
Bond angle in\(\mathrm{PH}_4^{+}\)is more than that of \(\mathrm{PH}_3\). This is because
3 hybridisation of P changes when \(\mathrm{PH}_3\) is converted to \(\mathrm{PH}_4^{+}\)
4 lone pair - bond pair repulsion exists in \(\mathrm{PH}_3\).
Explanation:
(d)
Both are \(s p^3\) hybridised. In \(\mathrm{PH}_4^{+}\), all the four orbitals are bonded whereas in \(\mathrm{PH}_3\) there is a lone pair of electrons on P , which is responsible for lone pairbond pair repulsion in \(\mathrm{PH}_3\) reducing the bond angle to less than \(109^{\circ} 28^{\prime}\).