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प्रश्न
Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
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उत्तर
The electronic configuration of fluorine is \[\ce{1s^2 2s^2 2p^5}\]. Thus it can form only one bond as it has only one unpaired electron. Electronic configuration of oxygen is \[\ce{1s^2 2s^2 2p^6 3s^2 3p^6}\].
It may be mentioned that oxygen also has vacant d-orbitals along with two 3p orbitals containing single electron. Thus, oxygen has greater bond formation capacity. In other words, it has greater ability to stabilize higher oxidation states.
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संबंधित प्रश्न
Why +2 oxidation state of manganese is more stable?
Account for the following:
Cr2+ is a strong reducing agent.
The \[\ce{E^{\circ}_{(M^{2+}/M)}}\] value for copper is positive (+0.34 V). What is possible reason for this? (Hint: consider its high ΔaH° and low ΔhydH°)
Use Hund’s rule to derive the electronic configuration of Ce3+ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Following are the transition metal ions of 3d series:
Ti4+, V2+, Mn3+, Cr3+
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
1) Which ion is most stable in an aqueous solution and why?
2) Which ion is a strong oxidising agent and why?
3) Which ion is colourless and why?
Give reasons for the following:
The transition metals generally form coloured compounds.
When \[\ce{KMnO4}\] solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because ______.
When a brown compound of manganese (A) is treated with \[\ce{HCl}\] it gives a gas (B). The gas taken in excess, reacts with \[\ce{NH3}\] to give an explosive compound (C). Identify compounds A, B and C.
A solution of \[\ce{KMnO4}\] on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
Match the properties given in Column I with the metals given in Column II.
| Column I (Property) | Column II (Metal) | |
| (i) | Element with highest second ionisation enthalpy |
(a) \[\ce{Co}\] |
| (ii) | Element with highest third ionisation enthalpy |
(b) \[\ce{Cr}\] |
| (iii) | \[\ce{M}\] in \[\ce{M(CO)6}\] is | (c) \[\ce{Cu}\] |
| (iv) | Element with highest heat of atomisation |
(d) \[\ce{Zn}\] |
| (e) \[\ce{Ni}\] |
Assertion: Separation of \[\ce{Zr}\] and \[\ce{Hf}\] is difficult.
Reason: Because \[\ce{Zr}\] and \[\ce{Hf}\] lie in the same group of the periodic table.
Answer the following question:
Which element of the first transition series has highest third ionisation enthalpy?
Identify the metal and justify your answer.
Carbonyl \[\ce{M(CO)5}\]
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The complex showing a spin-span magnetic moment of 2.82 B.M. is :-
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Give a reason for the following:
Zinc, cadmium and mercury are considered as d-block elements but not regarded as transition elements.
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29, 59, 74, 95, 102, 104
