What is the oxidation number of Co in [Co(NH₃)₄Cl(NO₂)]
What is the oxidation number of Co in [Co(NH₃)₄Cl(NO₂)]
Options
(a) 4 (b) 3 (c) 2 (d) 5
Correct Answer:
2
Explanation:
[Co(NH₃)₄Cl(NO₂)] → x + 4 * 0 + (-1) + (-1) […]
What is the oxidation number of Co in [Co(NH₃)₄Cl(NO₂)]
Options
(a) 4 (b) 3 (c) 2 (d) 5
Correct Answer:
2
Explanation:
[Co(NH₃)₄Cl(NO₂)] → x + 4 * 0 + (-1) + (-1) […]
Siver chloride is soluble in methylamine due to the formation of
Options
(a) [Ag(CH₃NH₂)₂]Cl (b) Ag(CH₃NH₂)Cl (c) AgOH (d) Ag + CH₃Cl + NH₄Cl
Correct Answer:
[Ag(CH₃NH₂)₂]Cl
Explanation:
Silver chloride dissolves in methyl amine […]
Which of the following is obtained when auric chloride reacts with sodium chloride
Options
(a) Na[AuCl] (b) Na[AuCl₂] (c) Na[AuCl₃] (d) Na[AuCl₄]
Correct Answer:
Na[AuCl₄]
Explanation:
No explanation available. Be the first to […]
Among the following the paramagnetic one is
Options
(a) Ni(CO)₄ (b) [Ni(CN)₄]² (c) [NiCl₄]² (d) [Ag(NH₃)₂]⁺
Correct Answer:
[NiCl₄]²
Explanation:
In[NiCl₄]²⁻, Cl⁻ provides a weak ligand field. Therefore it is unable to pair […]
The IUPAC name for Co(NH₃)₆Cl₃ is
Options
(a) cobalt nexaammine chloride (b) hexaammine cobalt (III) chloride (c) hexaammonia cobalt trichloride (d) hexaammine cobalt trichloride
Correct Answer:
hexaammonia cobalt trichloride
Explanation:
No explanation available. […]
Silver sulphide dissolves in a solution of sodium cyanide to form the complex
Options
(a) Na[Ag(CN)₂] (b) Na₃[Ag(CN)₄] (c) Na₅[Ag(CN)₆] (d) Na₂[Ag(CN)₂]
Correct Answer:
Na[Ag(CN)₂]
Explanation:
No explanation available. Be the first to […]
Which of the following carbonyls will have the strongest C – O bond
Options
(a) Mn(CO)₆⁺ (b) Cr(CO)₆ (c) V(CO)₆⁻ (d) Fe(CO)₅
Correct Answer:
Mn(CO)₆⁺
Explanation:
The presence of positive charge on the […]
Which one of the following complexes is not expected to exhibit isomerism
Options
(a) [Ni(NH₃)₄(H₂O)₂]²⁺ (b) [Pt(NH₃)₂Cl₂] (c) [Ni(NH₃)₂Cl₂] (d) [Ni(en)₃]²⁺
Correct Answer:
[Ni(NH₃)₂Cl₂]
Explanation:
No explanation available. Be the first to write […]
Platinum, palladium and iridium are called noble metals because
Options
(a) Alfred Noble discovered them (b) they are shining lustrous and pleasing to look at (c) they are found in native state (d) […]
In Cu-ammonia complex, the state of hybridization of Cu²⁺ is
Options
(a) sp³ (b) d³ (c) sp²f (d) dsp²
Correct Answer:
dsp²
Explanation:
Square planar complex is formed by dsp² hybridisation.
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