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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

A triple bond is made of :

  1. one $\sigma$- and two $\pi$ - bonds
  2. two $\sigma$- and one $\pi$ - bonds
  3. three $\sigma$- and three $\pi$ - bonds
  4. one $\sigma$- and four $\pi$ - bonds
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A triple bond is made of one $\sigma $-bond and two $\pi$-bonds.

So answer is A.

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

Number of $\sigma$ and $\pi$ bonds present in $CH _3-CH=CH- C \equiv CH$ are:

  1. $10\, \sigma,\, 3\, \pi$
  2. $10\, \sigma,\, 2\, \pi$
  3. $9\, \sigma,\, 2\, \pi$
  4. $8\, \sigma,\, 3\, \pi$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sigma bonds are formed by axial overlap and pi bonds are formed by sideways overlap. total number of bonds are $10\, \sigma,\, 3\, \pi$.

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 $N _2$ molecule, the atoms are bonded by:

  1. $1$ $\sigma$ and $2$ $\pi$ - bonds, $2$ L.P.
  2. $1$ $\sigma$ and $1$ $\pi$ - bonds, $1$ L.P.
  3. $2$ $\sigma$ and $1$ $\pi$ - bonds, No L.P.
  4. $1$ $\sigma$, $2$ $\pi$, and No L.P.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ N } _{ 2 }$ molecule contains $2$ lone pairs and $1s$ and $2p$ bonds.
$:N\equiv N:$
A triple bond consist of $1$ sigma and two $pi$ bonds. Lone pair refers to a pair of valence electrons that are not shared with another atom, it is also called non-bonding pair.
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 the correct statement (s) about the compound $NO[BF _4]$

  1. It has 5$\sigma$ and 2$\pi$ bond
  2. Nitrogen - oxygen bond length is higher than nitric oxide $(NO)$

  3. It is a diamagnetic species
  4. $BF$ bond length in this compound is lower than in $BF _3$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

$NO[BF _4]$ can be separated into $NO^+$ and $BF _4^-$. 
$BF _4^-$ has $4$  bonds, as single bonds between $B$ and $F$.
$NO^+$ has $1$  bonds, $2$ bonds.
It is a diamagnetic species.

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 the cyclo-$S _8$ molecule of rhombic sulphur, all the $S-S$ bond lengths and all the $S-S-S$ bond angles are respectively (give approximate values): 

  1. $204$ pm and $105^{\circ}$
  2. $102$ pm and $120^{\circ}$
  3. $204$ pm and $180^{\circ}$
  4. $102$ pm and $60^{\circ}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There is a single bond between each pair of $S$ atoms and two non-bonding electron pairs on each $S$ atom.Thus we see four electron domains around each $S$ atom and we would expect a tetrahedral geometry corresponding to $sp^3$ hybridisation which gives us the bond angle around $109^0$.

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 number of sigma and pi bonds in the structure:
$CH _2\, =\, CH\, -\, C\, \equiv\, CH$

  1. 7 and 3

  2. 6 and 2

  3. 4 and 3

  4. All are sigma bonds
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In $H _{2}C = CH-C\equiv CH$, there are $4: C-H$ sigma bonds and $3: C-C$ sigma bonds, thus it has a total of $7 \sigma$ bonds. Also there are $3 \pi$ bonds in the structure.

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 number of $\sigma$ bonds and $\pi$-bonds present in $\text{pent-4-en-1-yne}$ is:

  1. $10,3$
  2. $4,9$
  3. $3,10$
  4. $9,4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$HC\equiv C-{CH} _{2}-CH={CH} _{2}$


$\text{Pent-4-en-1-yne}$

Triple bond between $C$ and $C$ contains one $\sigma$ bond and two $\pi$ bonds. Double bond between $C$ and $C$ contains one $\sigma$ bond and one $\pi$ bond.

Total $\sigma$ bonds$=10$

Total $\pi$ bonds $=3$

Hence, the correct option is $\text{A}$

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

Identify the correct order of bond strengths?

  1. $\sigma _{p-p}>\pi _{p-p}>\sigma _{s-s}$
  2. $\sigma _{p-p}>\sigma _{s-s}>\pi _{p-p}$
  3. $\sigma _{s-s}>\sigma _{p-p}>\pi _{p-p}$
  4. $\pi _{p-s}>\sigma _{s-s}>\sigma _{p-p}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The correct order of bond strengths is $\displaystyle \sigma _{s-s}>\sigma _{p-p}>\pi _{p-p}$
The extent of overlap is higher for s orbital than for  p orbital. Higher is the extent of overlap, stronger is the bond formed.
Also, $\displaystyle \sigma$ bond is stronger than $\displaystyle \pi$ bond.