Discussion Forum



1)

 The pair(s)  of reagents that yield paramagnetic  species is/are


A) Na and excess of $ NH_{3}$

B) K and excess of $O_{2}$

C) Cu and dilute HN$O_{3}$

D) $O_{2}$ and 2-ethylanthraquinol

Answer:

Option A,B,C

Explanation:

Plan: the paramagnetic character of species can be easily explained on the basis of the presence of unpaired electrons. The compounds containing unpaired electron(s)  is/are paramagnetic.

 Reaction of alkali metals with ammonia depends upon the physical state of ammonia whether it is in gaseous state or a liquid state. If ammonia is considered as a gas then the reaction will be

  $(a) Na+NH_{3}\rightarrow NaNH_{3}+\frac{1}{2}H_{2}$

              (excess)

 (NaNH2+1/2 H2  are diamaganetic )  If ammonia  is considered as a liquid  then reaction will be

  $M+(x+y)NH_{3}\rightarrow[M(NH_{3})_{x}]^{+}+[e(NH_{3})_{y}]^{-}$

         . ammoniated electron

          . blue colour

           . paramagnetic

            . very strong reducing agent

(b)        $K+O_{2}\rightarrow KO_{2}(K^{+},O_2^-)$

                 (Excess)                      Potassium

                                           superoxide paramagnetic

  (c)   $3Cu+8HNO_{3}\rightarrow3Cu(NO_{3})_{2}+2NO+4H_{2}O$

                                     paramagnetic      paramagnetic     

  (d)

2232021928_ethy.JPG

 

Hence, option (a) , (b)  and (c) are correct choices.