Tag: tetravalence of carbon: shapes of organic compounds

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Multiple choice chemistry introduction to organic chemistry bonding in organic molecule tetravalence of carbon: shapes of organic compounds introduction to carbon and its compounds

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$\pi$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $C _3$.

Multiple choice chemistry introduction to organic chemistry bonding in organic molecule tetravalence of carbon: shapes of organic compounds introduction to carbon and its compounds

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. $C = C < C \equiv C < C - C$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $B$ 
${C \equiv C < C = C < C - C}$

Multiple choice chemistry introduction to organic chemistry bonding in organic molecule tetravalence of carbon: shapes of organic compounds introduction to carbon and its compounds

$Mg _{2}C _{3}$ 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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The C3(4-) ion is the propynide ion, which has the structure C-C=C. It contains two sigma bonds and two pi bonds.

Multiple choice chemistry introduction to organic chemistry bonding in organic molecule tetravalence of carbon: shapes of organic compounds introduction to carbon and its compounds

$\displaystyle:Mg _{2}C _{3}$ 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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice chemistry introduction to organic chemistry bonding in organic molecule tetravalence of carbon: shapes of organic compounds introduction to carbon and its compounds

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.
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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-} $

Multiple choice chemistry introduction to organic chemistry bonding in organic molecule tetravalence of carbon: shapes of organic compounds introduction to carbon and its compounds

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. $P _4S _{10}$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
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 ${ P } _{ 4 }{ S } _{ 10 }$

Multiple choice chemistry introduction to organic chemistry bonding in organic molecule tetravalence of carbon: shapes of organic compounds introduction to carbon and its compounds

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 $n$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $n$.

Multiple choice chemistry introduction to organic chemistry bonding in organic molecule tetravalence of carbon: shapes of organic compounds introduction to carbon and its compounds

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.