Tag: introduction to organic chemistry

Questions Related to introduction to organic chemistry

Pent -2-yne has how many $\sigma$ and $\pi$ bonds?

  1. 10 $\sigma$, 2$\pi$

  2. 12 $\sigma$, 2$\pi$

  3. 15 $\sigma$, 2$\pi$

  4. 13 $\sigma$, 3$\pi$


Correct Option: B
Explanation:

$CH _3-C\equiv C-CH _2-CH _3$
Pent-2-yne has 12$\sigma$ and 2$\pi$ bonds.
No. of $\sigma$ bonds = $C _1$ has 3 C-H bonds + $C _4$ has 2 C-H bonds + $C _5$ has 3 C-H bonds + 4 C-C bonds = 12$\sigma$ bonds.
No. of $\pi$ bonds = 2 bonds between $C _2$ and $C _3$.

The correct sequence of bond length in a single bond, a double bond and triple bond of $C$ is:

  1. ${(C - C) = (C = C) = (C \equiv C)}$

  2. ${C \equiv C < C = C < C - C}$

  3. $C - C < C = C < C \equiv C$

  4. $C = C < C \equiv C < C - C$


Correct Option: B
Explanation:

The bond length changes with the bond order like this:-

$Bond\ length\ \alpha \ \dfrac { 1 }{ Bond\ order } \ \alpha \ \dfrac { 1 }{ No.\ of\ bonds }$

As the number of bonds will increase, the bond order will increase and the bond length will decrease.
So, the correct option is $B$ 
${C \equiv C < C = C < C - C}$

$Mg _{2}C _{3}$ reacts with water forming propyne, $C _{2}^{4-}$ has

  1. Two sigma and two pi bonds

  2. Three sigma and one pi bonds

  3. Two sigma and one pi bonds

  4. Two sigma and three pi bonds


Correct Option: A

$\displaystyle:Mg _{2}C _{3}$ reacts with water forming propyne, $C _{3}^{4-}$ has:

  1. two sigma and two pi bonds

  2. three sigma and one pi bonds

  3. two sigma and one pi bonds

  4. two sigma and three pi bonds


Correct Option: A
Explanation:

Structure is: $^-C\equiv C-C^{-3}$
So it contains 2 sigma bond and 2 pi bonds.

Select correct statements about $C^{4-} _3$ ion.

  1. It reacts with $H _2O$ forming $C _3H _6$.

  2. It reacts with $H _2O$ forming $C _3H _4$.

  3. It has two sigma and two pi bonds.

  4. It has three sigma and one pi bonds.


Correct Option: B,C
Explanation:

The correct statements about $C^{4-} _3$ ion are

(B) It reacts with $H _2O$ forming $C _3H _4$

$\displaystyle  C^{4-} _3 + 2H _2O \rightarrow C _3H _4+O _2 + 4e^-$

(C) It has two sigma and two pi bonds $\displaystyle [C=C=C]^{4-} $

Which of the following molecule(s) has/have $d _n\, -\, p _{\pi}$ bonding?

  1. $H _2S _2O _3$

  2. $P _4O _{10}$

  3. $SO _3$

  4. $P _4S _{10}$


Correct Option: A,B,C,D
Explanation:

In ${ H } _{ 2 }{ S } _{ 2 }{ O } _{ 3 }$ $S$ empty $d$ orbital and $O$ has electrons in $P$ orbital.

Hence, it has ${ d } _{ n }$$-$${ P } _{ n }$ bonding.
In ${ P } _{ 4 }{ O } _{ 10 }$ $P$ have empty $d$ orbital and $O$ has electrons in $P$ orbital
Hence, it has ${ d } _{ n }$$-$${ P } _{ n }$ bonding.
In ${ SO } _{ 3 }$ $S$ have empty $d$ orbital and $O$ has electrons in $P$ orbital.
Hence, it has ${ d } _{ n }$$-$${ P } _{ n }$ orbital.
Similarly in ${ P } _{ 4 }{ S } _{ 10 }$

If "$n$" number of $H _3PO _4$ molecules are polymerized to produce chain molecule and ring molecule separately, then the number of $P - O - P$ linkages formed is, respectively :

  1. $n$ and $(n - 1)$

  2. $(n - 1)$ and $(n - 1)$

  3. $(n - 1)$ and $n$

  4. $n$ and $n$


Correct Option: C
Explanation:

In case of a ring, the number of linkages will be one less than the number of $P$ atoms because of the fact that each $P$ atom is shared by $2$ rings.
However, in case of a chain no sharing takes place and the number of linkages is equal to $n$.

In $Fe(CO) _5$, the $Fe-C$ bond possesses:

  1. $\pi$ - character only

  2. both $\sigma$ and $\pi$ - characters

  3. ionic character

  4. $\sigma$ - character only


Correct Option: B
Explanation:
It's called “synergic bonding”. The ligand (CO) donates it's lone pair of electrons to the vacant orbitals of the iron atom and forms the sigma-bond. Since the iron atom also possesses some electrons in it's d-orbitals, it back donates those electrons to the molecular orbitals of the ligand forming a π-bond. In this way, the metal-carbon bond length is reduced and the complex gets more stability. One important thing to keep in mind is that the metal atom donates it's electron pairs to the antibonding MO of CO, so the C-O bond is weakened by this synergic bonding, leading to a larger C-O bond length in the complex (as opposed to a free CO molecule).
Metal-C bond length reduces, C-O bond length increases, complex gets stability.
Hence, option $B$ is correct.