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प्रश्न
Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number.
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उत्तर
Formula of oxo-anion of manganese (Mn) in which it shows the oxidation state equal to its group number (7) is `Mn_4^(-)`
Mn + (-2 ×4) = -1
Mn - 8 = -1
Mn = +7
Oxidation of Mn in `MnO_4^(-)` is +7 which is equal to its group number 7.
संबंधित प्रश्न
Complete the following equations:
Complete the following equations : 2 MnO2 + 4 KOH + O2 →
Complete the following equations : 2 Na2CrO4 + 2 H + →
Give an example and suggest a reason for the following feature of the transition metal chemistry:
The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
Which of the following is amphoteric oxide?
\[\ce{Mn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4}\]
In the form of dichromate, \[\ce{Cr (VI)}\] is a strong oxidising agent in acidic medium but \[\ce{Mo (VI)}\] in \[\ce{MoO3}\] \[\ce{and W (VI)}\] in \[\ce{WO3}\] are not because:
(i) \[\ce{Cr(VI)}\] is more stable than \[\ce{Mo(VI)}\] and \[\ce{and W(VI)}\].
(ii) \[\ce{Mo(VI)}\] and \[\ce{and W(VI)}\] are more stable than \[\ce{Cr(VI)}\].
(iii) Higher oxidation states of heavier members of group-6 of transition series are more stable.
(iv) Lower oxidation states of heavier members of group-6 of transition series are more stable.
KMnO4 is coloured due to ______.
Indicate the steps in the preparation of \[\ce{K2Cr2O7}\] from chomite ore.
Indicate the steps in the preparation of \[\ce{K2Cr2O7}\] from chromite ore.
Indicate the steps in the preparation of \[\ce{K2Cr2O7}\] from chromite ore.
