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The electronic configuration of four elements are:
(I) [Kr]5s1
(II) [Rn]5f146d17s2
(III) [Ar]3d104s24p5
(IV) [Ar]3d64s2
(I) [Kr]5s1
(II) [Rn]5f146d17s2
(III) [Ar]3d104s24p5
(IV) [Ar]3d64s2
Consider the following statements:
(i) I shows variable oxidation state
(ii) II is a d-block element
(iii) The compound formed between I and III is covalent
(iv) IV shows single oxidation state
(i) I shows variable oxidation state
(ii) II is a d-block element
(iii) The compound formed between I and III is covalent
(iv) IV shows single oxidation state
Which statement is True(T) or Galse (F):
- a)FTFF
- b)FTFT
- c)FFTF
- d)FFFF
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The electronic configuration of four elements are:(I) [Kr]5s1(II) [Rn]...
1-it shows only +2 oxidation state- Hence False
2-is an f block element- Hence False
3-the bond formed is ionic because 1 is s block and ot is rb and 3 is a halide- Hence False
4-it is a d block element showing multiple oxidation states- Hence False
2-is an f block element- Hence False
3-the bond formed is ionic because 1 is s block and ot is rb and 3 is a halide- Hence False
4-it is a d block element showing multiple oxidation states- Hence False
Most Upvoted Answer
The electronic configuration of four elements are:(I) [Kr]5s1(II) [Rn]...
Electronic Configuration of Elements and Statements
Electronic Configuration:
(I) [Kr]5s1
(II) [Rn]5f146d17s2
(III) [Ar]3d104s24p5
(IV) [Ar]3d64s2
Statements:
(i) I shows variable oxidation state.
(ii) II is a d-block element.
(iii) The compound formed between I and III is covalent.
(iv) IV shows a single oxidation state.
Answer: FFFF
Explanation:
(i) I shows variable oxidation state: False
Element I has only one valence electron in its outermost shell, which is easily lost to form a cation with a +1 oxidation state. Hence it shows a fixed oxidation state.
(ii) II is a d-block element: False
Element II belongs to the f-block and is a member of the actinide series. It has electrons filling in the 5f orbitals.
(iii) The compound formed between I and III is covalent: False
Element I is a metal and Element III is a non-metal. Hence, the compound formed between them is expected to be ionic and not covalent.
(iv) IV shows a single oxidation state: False
Element IV belongs to the d-block and has electrons filling in the d-orbitals. As a result, it can exhibit variable oxidation states.
Electronic Configuration:
(I) [Kr]5s1
(II) [Rn]5f146d17s2
(III) [Ar]3d104s24p5
(IV) [Ar]3d64s2
Statements:
(i) I shows variable oxidation state.
(ii) II is a d-block element.
(iii) The compound formed between I and III is covalent.
(iv) IV shows a single oxidation state.
Answer: FFFF
Explanation:
(i) I shows variable oxidation state: False
Element I has only one valence electron in its outermost shell, which is easily lost to form a cation with a +1 oxidation state. Hence it shows a fixed oxidation state.
(ii) II is a d-block element: False
Element II belongs to the f-block and is a member of the actinide series. It has electrons filling in the 5f orbitals.
(iii) The compound formed between I and III is covalent: False
Element I is a metal and Element III is a non-metal. Hence, the compound formed between them is expected to be ionic and not covalent.
(iv) IV shows a single oxidation state: False
Element IV belongs to the d-block and has electrons filling in the d-orbitals. As a result, it can exhibit variable oxidation states.
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Community Answer
The electronic configuration of four elements are:(I) [Kr]5s1(II) [Rn]...
(1) first show+1 oxidation state because it belongs to alkali metal which gives +1 and -1(rare) oxidation state. (F)
(2) second is f - block elements (Lr) which belongs to Actinide series with 103 atomic no. (F)
(3) in this case first element is rubidium(metal) and second is bromine(non-metal) hence, it these makes ionic bond (as we know ionic bond formed by metal and non-metal). (F)
(4) in last case elements is iron which show+6 oxidation state. (F) hence, option 'd' is correct
(2) second is f - block elements (Lr) which belongs to Actinide series with 103 atomic no. (F)
(3) in this case first element is rubidium(metal) and second is bromine(non-metal) hence, it these makes ionic bond (as we know ionic bond formed by metal and non-metal). (F)
(4) in last case elements is iron which show+6 oxidation state. (F) hence, option 'd' is correct
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The electronic configuration of four elements are:(I) [Kr]5s1(II) [Rn]5f146d17s2(III) [Ar]3d104s24p5(IV) [Ar]3d64s2Consider the following statements:(i) I shows variable oxidation state(ii) II is a d-block element(iii) The compound formed between I and III is covalent(iv) IV shows single oxidation stateWhich statement is True(T) or Galse (F):a)FTFFb)FTFTc)FFTFd)FFFFCorrect answer is option 'D'. Can you explain this answer?
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Solutions for The electronic configuration of four elements are:(I) [Kr]5s1(II) [Rn]5f146d17s2(III) [Ar]3d104s24p5(IV) [Ar]3d64s2Consider the following statements:(i) I shows variable oxidation state(ii) II is a d-block element(iii) The compound formed between I and III is covalent(iv) IV shows single oxidation stateWhich statement is True(T) or Galse (F):a)FTFFb)FTFTc)FFTFd)FFFFCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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