In which of the given reaction the hybridization of the central metal atom
(underlined) is not changing?
Answers
Answer:
The numbers placed in front of formulas to balance equations are called coefficients, and they multiply all the atoms in a formula. Thus, the symbol “2 NaHCO3” indicates two units of sodium bicarbonate, which contain 2 Na atoms, 2 H atoms, 2 C atoms, and 6 O atoms (2 X 3= 6, the coefficient times the subscript for O)
Explanation:
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Answer:
Let us discuss each of the given reactions in detail.
(A)
P
H
3
+H
+
→PH
4
+
The hybridization of P remains the same, i.e., sp
3
hybridization.
In PH
3
, P atom contains 3 bond pairs of electrons and one lone pair of electrons.
In PH
4
+
, P atom contains 4 bond pairs of electrons.
(B)
B
F
3
+H
2
S→H
2
S→BF
3
The hybridization of B changes from sp
2
to sp
3
.
In BF
3
, B contains 3 bond pairs and zero lone pair of electrons.
In H
2
S→BF
3
, B contains 4 bond pairs of electrons and zero lone pair of electrons.
(C) H
3
B
O
3
→HBO
2
+H
2
O
The hybridization of B changes from sp
2
to sp.
In H
3
BO
3
, B contains three bond pairs of electrons and zero lone pairs of electrons.
In HBO
2
, B contains two bonding domains.
(D) 2H
Cl
O
2
→HClO+HClO
3
The hybridization changes from sp
3
to no hybridization.
In HClO
2
, Cl has 2 bonding domains and 2 lone pairs of electrons. It undergoes sp
3
hybridization.
In HClO
3
, Cl has 3 bond pairs of electrons and 1 lone pair of electrons and undergoes sp
3
hybridization.
Hence, the first and fourth reactions do not undergo any change in the hybridisation of the underlined atom.