NEET Test Series from KOTA - 10 Papers In MS WORD
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Chemical Bonding and Molecular Structure
12127
The number of shared pairs of electrons in propane is
1 \(2\)
2 \(4\)
3 \(6\)
4 \(10\)
Explanation:
(d) \(H - \mathop {\mathop C\limits_|^| }\limits_H^H - \mathop {\mathop C\limits_|^| }\limits_H^H - \mathop {\mathop C\limits_|^| }\limits_H^H - H\) There are \(10\) shared pairs of electrons.
Chemical Bonding and Molecular Structure
12014
Fluorine is more electronegative than either boron or phosphorus. What conclusion can be drawn from the fact that \(B{F_3}\) has no dipole moment but \(P{F_3}\) does
1 \(B{F_3}\) is not spherically symmetrical but \(P{F_3}\) is
2 \(B{F_3}\) molecule must be linear
3 The atomic radius of \(P\) is larger than the atomic radius of \(B\)
4 The \(B{F_3}\) molecule must be planar triangular
Explanation:
The \(BF_3\) molecule must be planar triangular. \(BF _3\) is planar triangular while \(PF _3\) is pyramidal.
Chemical Bonding and Molecular Structure
12150
The correct order towards bond angle is
1 \(sp < s{p^2} < s{p^3}\)
2 \(s{p^2} < sp < s{p^3}\)
3 \(s{p^3} < s{p^2} < sp\)
4 Bond angle does not depend on hybridisation
Explanation:
(c)Increasing order of bond angle is \(\mathop {s{p^3}}\limits_{109^o } < \mathop {s{p^2}}\limits_{120^o } < \mathop {sp}\limits_{180^o } \).
12127
The number of shared pairs of electrons in propane is
1 \(2\)
2 \(4\)
3 \(6\)
4 \(10\)
Explanation:
(d) \(H - \mathop {\mathop C\limits_|^| }\limits_H^H - \mathop {\mathop C\limits_|^| }\limits_H^H - \mathop {\mathop C\limits_|^| }\limits_H^H - H\) There are \(10\) shared pairs of electrons.
Chemical Bonding and Molecular Structure
12014
Fluorine is more electronegative than either boron or phosphorus. What conclusion can be drawn from the fact that \(B{F_3}\) has no dipole moment but \(P{F_3}\) does
1 \(B{F_3}\) is not spherically symmetrical but \(P{F_3}\) is
2 \(B{F_3}\) molecule must be linear
3 The atomic radius of \(P\) is larger than the atomic radius of \(B\)
4 The \(B{F_3}\) molecule must be planar triangular
Explanation:
The \(BF_3\) molecule must be planar triangular. \(BF _3\) is planar triangular while \(PF _3\) is pyramidal.
Chemical Bonding and Molecular Structure
12150
The correct order towards bond angle is
1 \(sp < s{p^2} < s{p^3}\)
2 \(s{p^2} < sp < s{p^3}\)
3 \(s{p^3} < s{p^2} < sp\)
4 Bond angle does not depend on hybridisation
Explanation:
(c)Increasing order of bond angle is \(\mathop {s{p^3}}\limits_{109^o } < \mathop {s{p^2}}\limits_{120^o } < \mathop {sp}\limits_{180^o } \).
12127
The number of shared pairs of electrons in propane is
1 \(2\)
2 \(4\)
3 \(6\)
4 \(10\)
Explanation:
(d) \(H - \mathop {\mathop C\limits_|^| }\limits_H^H - \mathop {\mathop C\limits_|^| }\limits_H^H - \mathop {\mathop C\limits_|^| }\limits_H^H - H\) There are \(10\) shared pairs of electrons.
Chemical Bonding and Molecular Structure
12014
Fluorine is more electronegative than either boron or phosphorus. What conclusion can be drawn from the fact that \(B{F_3}\) has no dipole moment but \(P{F_3}\) does
1 \(B{F_3}\) is not spherically symmetrical but \(P{F_3}\) is
2 \(B{F_3}\) molecule must be linear
3 The atomic radius of \(P\) is larger than the atomic radius of \(B\)
4 The \(B{F_3}\) molecule must be planar triangular
Explanation:
The \(BF_3\) molecule must be planar triangular. \(BF _3\) is planar triangular while \(PF _3\) is pyramidal.
Chemical Bonding and Molecular Structure
12150
The correct order towards bond angle is
1 \(sp < s{p^2} < s{p^3}\)
2 \(s{p^2} < sp < s{p^3}\)
3 \(s{p^3} < s{p^2} < sp\)
4 Bond angle does not depend on hybridisation
Explanation:
(c)Increasing order of bond angle is \(\mathop {s{p^3}}\limits_{109^o } < \mathop {s{p^2}}\limits_{120^o } < \mathop {sp}\limits_{180^o } \).
12127
The number of shared pairs of electrons in propane is
1 \(2\)
2 \(4\)
3 \(6\)
4 \(10\)
Explanation:
(d) \(H - \mathop {\mathop C\limits_|^| }\limits_H^H - \mathop {\mathop C\limits_|^| }\limits_H^H - \mathop {\mathop C\limits_|^| }\limits_H^H - H\) There are \(10\) shared pairs of electrons.
Chemical Bonding and Molecular Structure
12014
Fluorine is more electronegative than either boron or phosphorus. What conclusion can be drawn from the fact that \(B{F_3}\) has no dipole moment but \(P{F_3}\) does
1 \(B{F_3}\) is not spherically symmetrical but \(P{F_3}\) is
2 \(B{F_3}\) molecule must be linear
3 The atomic radius of \(P\) is larger than the atomic radius of \(B\)
4 The \(B{F_3}\) molecule must be planar triangular
Explanation:
The \(BF_3\) molecule must be planar triangular. \(BF _3\) is planar triangular while \(PF _3\) is pyramidal.
Chemical Bonding and Molecular Structure
12150
The correct order towards bond angle is
1 \(sp < s{p^2} < s{p^3}\)
2 \(s{p^2} < sp < s{p^3}\)
3 \(s{p^3} < s{p^2} < sp\)
4 Bond angle does not depend on hybridisation
Explanation:
(c)Increasing order of bond angle is \(\mathop {s{p^3}}\limits_{109^o } < \mathop {s{p^2}}\limits_{120^o } < \mathop {sp}\limits_{180^o } \).
