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6 months ago

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P
Payal Gupta

Contributor-Level 10

A. [ F e F 6 ] 3           

Fe3+ - 3d54s0

F- is weak field ligand, so no pairing of electrons :

                      

Number of unpaired electrons, n = 5

B. [ F e ( C N ) 6 ] 3          

F e 3 + 3 d 5 4 s 0                              

CN- is strong field ligand, so pairing of electrons takes place.

                            

Number of unpaired electrons, n = 1

C. [ M n C l 6 ] 3          

...more

New answer posted

6 months ago

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P
Payal Gupta

Contributor-Level 10

Electronic configuration of the given ions is :

E u 2 + 4 f 7 6 s 0                 

T m 2 + 4 f 1 3 6 s 0 S m 2 + 4 f 6 6 s 0 T m 3 + 4 f 1 2 6 s 0

T b 4 + 4 f 7 6 s 0 Y b 2 + 4 f 1 4 6 s 0 D y 3 + 4 f 9 6 s 0 y b 3 + 4 f 1 3 6 s 0              

Hence, Eu2+ and Tb4+ have half | filled f-orbitals and Yb2+ have completely filled f-orbitals.

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6 months ago

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Payal Gupta

Contributor-Level 10

In 3d-series, all metals except Cu have negative value of E M 2 + / M 0 due to low hydration enthalpy and high I.E and sublimation enthalpy.

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6 months ago

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Payal Gupta

Contributor-Level 10

Fluorine forms only one oxoacid which is hypofluorous acid HOF because it shows only – 1 oxidation state. Which is due to its the smallest size among halogens & the highest electronegativity.

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6 months ago

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Payal Gupta

Contributor-Level 10

N H 4 C l ( a q ) + N a N O 2 ( a q ) N 2 ( g ) + 2 H 2 O ( l ) + N a C l ( a q )  

Hence, N2 gas is produced.

New answer posted

6 months ago

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P
Payal Gupta

Contributor-Level 10

For isoelectronic species

I o n i c r a d i i N u m b e r o f e N u m b e r o f p              

N a + M g 2 + N 3 O 2 F e = 1 0 1 0 1 0 1 0 1 0 p = 1 1 1 2 7 8 9                

Hence, correct order of ionic radii is;

M g 2 + < N a + < F < O 2 < N 3                

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6 months ago

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Payal Gupta

Contributor-Level 10

Higher the E.N. difference between hydrogen and other atom then higher be the strength of intermolecular H-bond

Here, order of difference in E.N is

O - H > N – H > C - H

Hence, correct order of H bond strength is,

CH4 < HCN < NH3

New answer posted

6 months ago

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Payal Gupta

Contributor-Level 10

Leaching involves the given reaction,

4 A u ( s ) + 8 C N ( a q ) + 2 H 2 O ( a q ) + O 2 ( g ) 4 [ A u ( C N ) 2 ] ( a q ) + 4 O H ( a q )              

Here, O2 is required for formation of Au (l) cyanide complex but no complex in absence of O2.

2 [ A u ( C N ) 2 ] ( a q ) + Z n ( s ) [ Z n ( C N ) 4 ] 2 ( a q ) + 2 A u ( s )                

In above displacement reaction, Zn is oxidized.

New answer posted

6 months ago

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Payal Gupta

Contributor-Level 10

Structure PCl5 is trigonal bipyramidal,

Hybridization of P is sp3d

Equatorial bonds lie in a plane

Axial bonds are longer than equatorial bonds so axial bonds are weaker than equatorial bonds.

New answer posted

6 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Fluorine forms only one oxoacid which is hypofluorous acid HOF because it shows only – 1 oxidation state. Which is due to its the smallest size among halogens & the highest electronegativity.

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