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miss Akunina [59]
4 years ago
7

How can u detect the which d-orbital is used in making bond ??

Chemistry
1 answer:
zavuch27 [327]4 years ago
4 0
This is much more simple than it sounds... 

s has 1 orbital; p = 3 orbitals; d= 5 orbitals; f= 7 orbitals. (you just need to memorize this) 
A maximum of 2 electrons can occupy each orbital. 

<span>The number of orbitals that each atom has is based on the number of electrons it has and by consequence it's position on the periodic table. </span>

The orbitals occur in sequence. Whereby electrons fill first from the lowest energy level (1s) outwards to the highest. 
3p = the following sequence. 
1s, 2s, 2p, 3s, 3p: these 'sets' can hold the following electrons respectively (2+2+6+2+6) 18 which corresponds with argon on the periodic table. REMEMBER p has 3 orbitals, d has 5 orbitals. So, here there are 9 orbitals. 

The sequence through n=4 is: 

<span>1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f 
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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3 years ago
Consider this equation: 2.524 g (5.1 × 106 g) ÷ (6.85 × 103 g) = ? How many significant figures should the result have?
weqwewe [10]

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7

Explanation:

2.524g(5.1)(106)g

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(HELP) What were the main themes of the Arab poems?​
Mazyrski [523]

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3 years ago
How does the periodic table indicate the number of valence electrons element has?
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7 0
3 years ago
What mass of iron (III) nitrate will be in 129.8ml of a 0.3556 molar aq iron (III) nitrate solution?
Anvisha [2.4K]

<u>Answer:</u> The mass of iron (III) nitrate is 11.16 g/mol

<u>Explanation:</u>

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 0.3556 M

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Volume of solution = 129.8 mL

Putting values in above equation, we get:

0.3556M=\frac{\text{Mass of iron (III) nitrate}\times 1000}{241.86 g/mol\times 129.8}\\\\\text{Mass of iron (III) nitrate}=11.16g

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