TS EAMCET 2019 :: Chemistry :: General questions

• Chemistry - General questions

Identify the oxidation  states of Mn, when $MnO_{4}^{2-}$ ion undergoes disproportionation reaction under acidic medium

Answer:

Explanation:

 The disproportionation reaction for $MnO_{4}^{2-}$ occurs as follows:

 $3 Mn^{+6}O_{4}^{2-}+4H^{+}\rightarrow Mn^{+4}O_{2}+2Mn^{+7}O_{4}^{-}+2H_{2}O$

 Hence, $MnO_{4}^{2-}$ is oxidised to $MnO_{4}^{-}$ and reduced to  $MnO_{2}$ . Thus  , oxidation  state of Mn when $MnO_{4}^{2-}$  undergoes disproportionation reaction under acidic medium are +7 and +4 

$\therefore$   Hence option (d)  is the correct answer.

What is the correction term in the pressure for real gas in comparison to the ideal gas?

Answer:

Explanation:

 According to van der Waal's correction for pressure for real -gases , the pressure correction is given by  

 $p+ \frac{an^{2}}{V^{2}}$.

 $\therefore$  Correction  term for pressure of a real gas =$\frac{an^{2}}{V^{2}}$

 and option (c) is the correct answer

How many of the following species are diamagnetic?

$He_{2}^{+}, H_{2}^{+}, H_{2}^{-}, He$

Answer:

Explanation:

 According to molecular orbital theory  , species with all paired electrons  in their electronic configuration are diamagnetic in nature

$He^{+}_{2}\rightarrow $  Has total 3 electrons , thus it contains one unpaired electron and is paramagnetic in nature.

$H_{2}\rightarrow $ Has total 2 electrons , thus it contains all paired electrons and is diamagnetic in nature

$H^{-}_{2}\rightarrow $ Has only 1 electron, thus it is paramagnetic in nature

$H^{2}_{2}\rightarrow $ Has total 3 electrons, thus it contains  one unpaired electron and is paramagnetic in nature

$He \rightarrow $  Contains total two (2) electrons and has no unpaired electron, thus is diamagnetic in nature.

 Hence , among the given species, we have only two of them which are diamagnetic and thus, option (b) is the correct answer.

Match the following.

Answer:

Explanation:

 (A)  Nodes (III)   $|\psi^{2}|$ is zero because $|\psi^{2}|$ represents the region where probability of electron finding is zero

(B)   Subsidiary quantum Number

 (I) is a three-dimensional shape of the orbital. It is also called the azimuthal quantum number (l). The values of (l) give three-dimensional shapes of orbitals

 (C)   White Light

 (V)  Continuous spectrum because when all the energy waves are so close to each other that they appear as in a continuous pattern, the spectrum is called a continuous spectrum

 (D) (II) Heisenberg's Uncertainty principle

 Significant only for motion of  microscopic objects, because , according to this principle, it is impossible to determine simultaneously , the exact position and exact momentum  (or velocity)  of an electron. Hence , (d)  is the correct option

From the following energy levels of hydrogen atom, the value of $E_{\infty}$ and $E_{3}$ in J are, respectively

______________   $E_{\infty}$`=-------------

______________ $E_{3}$ =----------------

______________ $E_{2}=-0.545 \times  10^{-18} J$

_____________  $E_{1}=-2.18 \times 10^{-18} J$

Answer:

Explanation:

$ E_{n}=-2.18 \times 10^{-18}. \frac {Z^{2}}{n^{2}}$

where, E_n  = energy of an electron  in nth orbital.

        Z= atomic number

      n= number of orbital

 For hydrogen , Z=1

 Therefore,

$E_{3}=-2.18\times 10^{-18}\times\frac{1}{3^{2}}$

 $=-\frac{2.18\times 10^{-18}}{9}$

 =$-0.242 \times 10^{-18}$

 and    $E_{\infty}=-2.18 \times 10^{-18}\times \frac{1}{\infty}$

   $E_{\infty}=0$

 Hence, $E_{\infty}$=0   and $E_{3}=-0.242 \times 10^{-18}$

 and option (c) is the correct answer.

• Chemistry - General questions