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Nezavi [6.7K]
3 years ago
8

The filling of which orbital is represented by the transition metals in period 4? 3d 3s 2s 2p

Chemistry
2 answers:
Natali5045456 [20]3 years ago
8 0

Answer:

3d

Explanation:

Transition metal are elements having partially filled d sub shell. In these metals last electron is filled in d sub shell.

General electronic configuration of transition metal: (n-1)^{1-10}ns^{1-2}

Where, n = outermost shell.

Period 4 means, last shell no. is 4.

For transition metals of period 4,

General electronic configuration: 3d^{1-10}4s^{1-2}

for example, Fe belong to periods 4.

Electronic configuration of Fe: [Ar]3d^64s^2

So, filling of 3d orbital is represented by transition metal of 4.

pogonyaev3 years ago
3 0
Transition metals are the metal located in the center block of the periodic table. Also, this block represents the D sublevel.

So, the answer is 3d.

Also, keep in mind that the D block starts "one period late". this means that if you are in period 4, the 3d orbital starts. if you are in period 5, that where 4d starts.
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Answer:

1.882 g

Explanation:

Data Given

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Reaction Given:

                       CH₄+ 4Cl₂ --------→ CCl₄ + HCl

Solution:

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                    1 mol        4 mol

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Now

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Molar mass of  Cl₂  = 71 g/mol

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                mass of Cl₂ = 4 mol x 71 g/mol

                mass of Cl₂  = 284 g

Molar mass of CH₄= 12+ 4(1)

Molar mass of CH₄= 16 g/mol

mass of CH₄

                mass in grams = no. of moles x molar mass

                mass of CH₄= 1 mol x 16 g/mol

                mass of CH₄ = 16 g

So,

284 g of Cl₂  combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂  

Apply unity Formula

                           284 g of Cl₂  ≅ 16 g of methane ( CH₄ )

                           33.4 g of Cl₂  ≅ X g of methane ( CH₄ )

By cross multiplication

                          X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

                          X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

So

methane (CH₄) = 1.882 g

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