JEE 2012 :: General Questions :: Chemistry

• General Questions - Physics
• General Questions - Chemistry
• General Questions - Mathematics

In allene ($C_{3}H_{4}$), the type(s) of hybridization of the carbon atoms, is(are)

Answer:

Explanation:

Allete is the name given to propdiene,

$H_{2}C=C=CH_{2}$

The hybridization of an atom is determined by determining the number of hybrids orbitals ar that atom which is equal to the number of sigma ($\sigma$)  bonds plus a number of lone pairs at the concerned atom.

Here, the terminal carbons have only three sigma bonds associated with them, therefore, hybridization of terminal carbons is sp³. The central carbon has only two sigma bonds associated, hence hybridization at central carbon is sp

For one mole of a van der Waals' gas when b = 0 and T = 300 K, the pV vs. 1/V plot is shown below. The value of the van der Waals' constant a (atm L mol^-2)

Answer:

Explanation:

The van der Waals' equation of state is

$\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)=RT$

for 1 mole and when b=0 so,

$\Rightarrow$    $pV=RT-\frac{a}{V}$ ..........(i)

Equation (i) is a straight equation between pV and $\frac{1}{V}$ whose slope is "-a'.

Equationg with slope of the straight line given in the graph 

   $-a=\frac{20.1-21.6}{3-2}=-1.5$

$\Rightarrow$ a=1.5

The number of aldol reaction(s) that   occurs in the given transformation is

Answer:

Explanation:

The given reaction is an example of repeated aldol condensation followed
by Cannizzaro reaction

Step I,

In the last step, formaldehyde is oxidize and the other aldehyde is reduced giving the desired products

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [$a_{0}$, is Bohr radius]

Answer:

Explanation:

According to Bohr's model

  $mvr=\frac{nh}{2\pi|}$

 $\Rightarrow$   $(mv)^{2}=\frac{n^{2}h^{2}}{4\pi^{2}r^{2}}$

$\Rightarrow$ $KE=\frac{1}{2}mv^{2}=\frac{n^{2}h^{2}}{8\pi^{2}r^{2}m}$......(i)

Also Bohr's radius for  H-atom is $r=n^{2}a_{0}$

Substiuting 'r' in equation (i) gives

 $KE= \frac{h^{2}}{8 \pi^{2}n^{2}a_{0}^{2}m}$

wher n=2,

  KE=$\frac{h^{2}}{32 \pi^{2}m a_{0}^{2}}$

Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?

Answer:

Explanation:

Let oxidation number of "N" be x

In $HNO_{3}$+1+x+3(-2)=0

$\Rightarrow$   x=+5

 In $NO$      x-2=0 $\Rightarrow$x=+2

 in $N_{2}$   x=0

 In $NH_{4}Cl$  x+4-1=0$\Rightarrow$ x=-3

• General Questions - Physics
• General Questions - Chemistry
• General Questions - Mathematics