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JEE Advanced 2017 Paper 1
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© examsnet.com
Question : 26
Total: 54
The sum of the number of lone pair of electrons on each central atom in the following species is
[
T
e
B
r
6
]
2
−
,
[
B
r
F
2
]
2
+
,
S
N
F
3
and
[
X
e
F
3
]
−
(Atomic numbers :N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54)
Your Answer:
Validate
Solution:
Number of electrons around central atom (N)
=
V
+
M
±
C
2
[V=Valence electrons of central atom,M=+1 for every monovalent atom,-1 for every trivalent atom,C=cationic or anionic charge]
[
T
e
B
r
6
]
2
−
:
N
=
6
+
6
+
2
2
=
7
Number of bp = 6
Number of lp = 1
[
B
r
F
2
]
+
:
N
=
7
+
2
−
1
2
=
4
Number of bp = 2
Number of lp = 2
$
SNF_3:N={6-1+3}/2=4$
Number of bp = 4
Number of lp = 0
[
X
e
F
3
]
−
:
N
=
8
+
3
+
1
2
=
6
Number of bp = 3
Number of lp = 3
The sum of the number of lone pairs of electrons on each central atom = 1 + 2 + 0 + 3 = 6
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