Question

Element A has 3 electrons in the outermost orbit and
element B has 6 electrons in the outermost orbit. The
formula of the compound formed between A and B
would be


nO SPaM​

Answer 1

Answer:

$\large\bold\red{A_{2}B_{3}}$

Explanation:

Given that,

Element A has 3 electrons in the outermost orbit.

Clearly, it's a Metal

Thus,

it will use it's all 3 electrons in forming the bond.

Therefore,

we have it's ionic form as $\bold{{A}^{3+}}$

Similarly,

Element B has 6 electrons in the outermost orbit.

Clearly, it's a non - metal .

Therefore,

it will need (8-6) i.e., 2 electrons to form the bond.

Thus,

it's ionic form will be $\bold{{B}^{3-}}$

Now,

From criss - cross method,

we know that,

2 atoms of A and 3 atoms of B will be required to form the bond.

Hence,

the molecular formula of the compound is$\bold{A_{2}B_{3}}$

Answer 2

Given that,Element A has 3 electrons in the outermost orbit.Clearly, it's a MetalThus,it will use it's all 3 electrons in forming the bond.Therefore,we have it's ionic form as \bold{{A}^{3+}}A 3+ Similarly,Element B has 6 electrons in the outermost orbit.

Given that,Element A has 3 electrons in the outermost orbit.Clearly, it's a MetalThus,it will use it's all 3 electrons in forming the bond.Therefore,we have it's ionic form as \bold{{A}^{3+}}A 3+ Similarly,Element B has 6 electrons in the outermost orbit.Clearly, it's a non - metal .

Given that,Element A has 3 electrons in the outermost orbit.Clearly, it's a MetalThus,it will use it's all 3 electrons in forming the bond.Therefore,we have it's ionic form as \bold{{A}^{3+}}A 3+ Similarly,Element B has 6 electrons in the outermost orbit.Clearly, it's a non - metal .Therefore,

Given that,Element A has 3 electrons in the outermost orbit.Clearly, it's a MetalThus,it will use it's all 3 electrons in forming the bond.Therefore,we have it's ionic form as \bold{{A}^{3+}}A 3+ Similarly,Element B has 6 electrons in the outermost orbit.Clearly, it's a non - metal .Therefore,it will need (8-6) i.e., 2 electrons to form the bond.

it's ionic form will be \bold{{B}^{3-}}B

it's ionic form will be \bold{{B}^{3-}}B 3−

it's ionic form will be \bold{{B}^{3-}}B 3−

it's ionic form will be \bold{{B}^{3-}}B 3− Now,

it's ionic form will be \bold{{B}^{3-}}B 3− Now,From criss - cross method,

it's ionic form will be \bold{{B}^{3-}}B 3− Now,From criss - cross method,we know that,

it's ionic form will be \bold{{B}^{3-}}B 3− Now,From criss - cross method,we know that,2 atoms of A and 3 atoms of B will be required to form the bond.

it's ionic form will be \bold{{B}^{3-}}B 3− Now,From criss - cross method,we know that,2 atoms of A and 3 atoms of B will be required to form the bond.Hence,

it's ionic form will be \bold{{B}^{3-}}B 3− Now,From criss - cross method,we know that,2 atoms of A and 3 atoms of B will be required to form the bond.Hence,the molecular formula of the compound is\bold{A_{2}B_{3}}A

it's ionic form will be \bold{{B}^{3-}}B 3− Now,From criss - cross method,we know that,2 atoms of A and 3 atoms of B will be required to form the bond.Hence,the molecular formula of the compound is\bold{A_{2}B_{3}}A 2

it's ionic form will be \bold{{B}^{3-}}B 3− Now,From criss - cross method,we know that,2 atoms of A and 3 atoms of B will be required to form the bond.Hence,the molecular formula of the compound is\bold{A_{2}B_{3}}A 2

it's ionic form will be \bold{{B}^{3-}}B 3− Now,From criss - cross method,we know that,2 atoms of A and 3 atoms of B will be required to form the bond.Hence,the molecular formula of the compound is\bold{A_{2}B_{3}}A 2 B

it's ionic form will be \bold{{B}^{3-}}B 3− Now,From criss - cross

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