MCQ
Predict the formula of a compound formed between a metal 'M' which has 1st, 2nd, 3rd IP values as 518, 7314, 9820KJmol−1 respectively and a halogen 'X'.
- AMX2
- BM2X3
- CMX3
- DMX
Explanation:
Alkali metals have low first ionization energies and relatively higher second and third ionization energies because the electron have to be remover from a stable octet.
As in the given question, there is more difference in the values of first and second ionization energies i.e. the metal has 1 valence electron,
∴ formula of the compound will be MX.
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$SO{{_{3}^{2-}}_{\left( aq. \right)}}+{{H}_{2}}{{O}_{\left( l \right)}}\to SO{{_{4}^{2-}}_{\left( aq. \right)}}+2{{H}^{+}}_{\left( aq. \right)}+2{{H}^{+}}_{\left( aq. \right)}+2{{e}^{-}}$
If the oxidation number of metal in the salt was $+3$, what would be the new oxidation number of metal?
$(I)$ $CH_3COO^{\Theta}$ $(II)$ $CH_3O^{\Theta}$
$(III)$ $CH_3CH_2^{\Theta}$ $(IV)$ $CH_3SO_3^{\Theta}$