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luda_lava [24]
3 years ago
9

Which of the following atoms could have an expanded octet when it is the central atom in a covalent compound?a) B

Chemistry
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

only chlorine can expand its octet.

Explanation:

An atom can expand its octet is it has empty d orbital

the electronic configuration of given elements will be:

B : 1s2 2s2 2p1 [Valence shell n =2 no d orbital]

O :1s2 2s2 2p4 [Valence shell n =2 no d orbital]

F : 1s2 2s2 2p5 [Valence shell n =2 no d orbital]

Cl :1s2 2s2 2p6 3s2 3p5 3d0 [Valence shell n =2 no d orbital]

Out of given elements only chlorine has empty d orbitals in its valence shell

Thus only chlorine can expand its octet.

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4 0
3 years ago
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What happens when The vapour obtained by dropping conc. H2SO4 in a mixture of KI and MnO2 is treated with hypo solution​
Shalnov [3]

Iodine is decolorized.

The first reaction stated in the question occurs as follows;

2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)

The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.

Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.

The equation of the titration reaction is;

2Na2S2O3 + I2→ Na2S4O6 + 2NaI

When this reaction takes place, iodine is decolorized due to its reduction to I^-.

6 0
3 years ago
Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO;?
faltersainse [42]

Answer:

K₂CO₃    

Explanation:

Given parameters:

Number of moles of K = 0.104mol

Number of moles of C = 0.052mol

Number of moles of O = 0.156mol

Method

From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.

Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.

  • Divide the given moles through by the smallest which is C, 0.052mol.
  • Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
  • This is the empirical formula

Solution

Elements                             K                       C                    O

Number of moles            0.104                0.052            0.156

Dividing by the

smallest                       0.104/0.052     0.052/0.052  0.156/0.052

                                            2                           1                     3

The empirical formula is K₂CO₃      

3 0
3 years ago
Does a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? Write the n
olganol [36]

Answer : Yes, a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

CaCl_2(aq)+2HgNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+Hg_2Cl_2(s)

The ionic equation in separated aqueous solution will be,

Ca^{2+}(aq)+2Cl^{-}(aq)+2Hg^{+}(aq)+2NO_3^{-}(aq)\rightarrow Hg_2Cl_2(s)+Ca^{2+}(aq)+2NO_3^{-}(aq)

In this equation, Ca^{2+}\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

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