Question

which elements do the following electron configurations represent: 66. 1s²2s²2p⁶3s²3p⁶ 67. 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁵ 68. 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹4d¹⁰5p² 69. ne3s²3p³ 70. kr5s²4d¹⁰ write the unabbreviated electron configuration for the following elements. 71. si 72. titanium 73. as write the nobel gas abbreviation for the following elements. 74. vanadium 75. iodine

Explanation:

Step1: Count electrons in 66

Add up the electrons in $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$. $2 + 2+6 + 2+6=18$. Element with 18 electrons is Argon (Ar).

Step2: Count electrons in 67

Add up electrons in $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{5}$. $2+2 + 6+2+6+2 + 10+5 = 35$. Element with 35 electrons is Bromine (Br).

Step3: Count electrons in 68

Add up electrons in $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{2}$. $2+2+6 + 2+6+2+10+6+2+10+2 = 50$. Element with 50 electrons is Tin (Sn).

Step4: Expand [Ne] in 69

[Ne] represents $1s^{2}2s^{2}2p^{6}$. So $[Ne]3s^{2}3p^{3}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$. Total electrons: $2+2+6+2+3 = 15$. Element with 15 electrons is Phosphorus (P).

Step5: Expand [Kr] in 70

[Kr] represents $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}$. So $[Kr]5s^{2}4d^{10}$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}$. Total electrons: $2+2+6+2+6+2+10+6+2+10 = 48$. Element with 48 electrons is Cadmium (Cd).

Step6: Un - abbreviate Si (71)

Silicon (Si) has atomic number 14. Electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}$.

Step7: Un - abbreviate Titanium (72)

Titanium (Ti) has atomic number 22. Electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{2}$.

Step8: Un - abbreviate As (73)

Arsenic (As) has atomic number 33. Electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3}$.

Step9: Nobel gas abbreviation for Vanadium (74)

Vanadium (V) has atomic number 23. Nobel gas before V is Ar. Remaining electrons: $23 - 18=5$. So the Nobel - gas abbreviation is $[Ar]4s^{2}3d^{3}$.

Step10: Nobel gas abbreviation for Iodine (75)

Iodine (I) has atomic number 53. Nobel gas before I is Kr. Remaining electrons: $53 - 36 = 17$. So the Nobel - gas abbreviation is $[Kr]5s^{2}4d^{10}5p^{5}$.

Answer:

1. Ar
2. Br
3. Sn
4. P
5. Cd
6. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{2}$
7. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{2}$
8. $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{3}$
9. $[Ar]4s^{2}3d^{3}$
10. $[Kr]5s^{2}4d^{10}5p^{5}$