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harina [27]
1 year ago
11

Nearly every compound of silicon has the element in the ⁺4 oxidation state. In contrast, most compounds of lead have the element

in the ⁺2 state.(c) Give an analogous example from Group 3A(13).
Chemistry
1 answer:
baherus [9]1 year ago
4 0

Boron shows analogous behaviour with silicon. Every compound of Boron have +3 oxidation state, while as we go down the the oxidation state become +1 and +3. For example, gallium, indium, etc.

Similarly, every compound of silicon has the element in the ⁺4 oxidation state. In contrast, most compounds of lead have the element in the ⁺2 state because of inert pair effect.

<h3>What is Inert pair effect? </h3>

The inert-pair effect is defined as the tendency of two electrons in the outermost atomic s-orbital almost remain unshared in the compounds of the post-transition metals.

<h3>How we calculate Oxidation state? </h3>
  • Each atom in an element either be in its uncombined or free state has oxidation number of zero. Such as each atom in H₂, Cl₂ , P4, ,O₂ , Na, Al, O3, S8, and Mg, all have an oxidation number zero.
  • The oxidation state of ions that comprise of only one atom is the actual charge on the ion.
  • The oxidation state of hydrogen is +1, excluding when it is bonded to metals having two elements. For example, CaH2, its oxidation state is –1.
  • Fluorine and other halogens have an oxidation state equal to –1 when they appear as a form of halide ions in their compounds.

Since the inert pair effect increases as we go down the group and become more predominant, therefore, the stability of +2 oxidation state goes on increasing down the group. Therefore, gallium, indium are mostly found in +1 oxidation state.

Thus, we concluded that Boron shows analogous behaviour with silicon. Every compound of Boron have +3 oxidation state, while as we go down the the oxidation state become +1 and +3. For example, gallium, indium, etc.

learn more about oxidation state:

brainly.com/question/25551544

#SPJ4

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Calculează masa zaharului si volumul apei necesare pentru prepararea 500g de soluție de zahăr cu partea de masa 20%
Sedbober [7]
T = 20 %  : 20 / 100 = 0.2

m1 = solute 

m2 = Solvent

T = m1 / m1 + m2

0.2 = 500 g / 500 g + m2

0.2 * ( 500 + m2 ) = 500

0.2 * 500 + 0.2 m2 = 500

100 + 0.2 m2 = 500

0.2 m2 = 500 - 100

0.2 m2 = 400

m2 = 400 / 0.2

m2 = 2000 g of water

hope this helps!



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