MCQ
Which of the following is paramagnetic ?
  • A
    $NO^+$
  • B
    $CO$
  • C
    $O_2^-$
  • $CN^-$

Answer

Correct option: D.
$CN^-$
d
Paramagnetism refers to the magnetic state of an atom with one or more unpaired electrons.

$O _{2}^{-}$ consists of one unpaired electrons.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Which one of the gas dissolves in ${H_2}S{O_4}$ to give oleum
The name of equation showing relation between electrode potential $(E)$standard electrode potential $({E^o})$ and concentration of ions in solution is
The rate constant of a first order reaction whose half-life is $ 480\,\sec$ , is
According to the first law of Faraday, the weight of a substance discharge at the electrode is
$\begin{matrix}
   \,\,\,\,\,C{{H}_{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   |\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   C{{H}_{3}}-C=C{{H}_{2}}\xrightarrow[{{R}_{2}}{{O}_{2}}]{HBr}X\xrightarrow{C{{H}_{3}}ONa}Y;  \\
\end{matrix}$ $X$ and $Y$ are
$C{H_3} - C \equiv C - C{H_3}\xrightarrow[\Delta ]{{{H^ \oplus }/KMn{O_4}}}A\xrightarrow[{(2)\,{H^ \oplus }}]{{(1)\,LiAl{H_4}}}B$

$B$ will be

Given below are two statements one is labelled as Assertion $A$ and the other is labelled as Reason $R:$

Assertion $A :$ The amphoteric nature of water is explained by using Lewis acid/base concept.

Reason $R$ : Water acts as an acid with $NH _{3}$ and as a base with $H _{2} S$.

In the light of the above statements choose the correct answer from the options given below :

$C$ will be
The correct sequence which shows decreasing order of the ionic radii of the elements is
What is the dissociation constant for $NH_4OH$ if at a given temperature its $0.1\,N$ solution has $pH = 11.27$ and the ionic product of water is $7.1 \times 10^{-15}$ (antilog $0.73 = 5.37$ )