Question

Okay so I am just having trouble with what element this would be?

Answer 1

should be ar-senic (this website will filter it out if you don't put the dash)
it has a full d-block
see what you can get from this

Answer 2

Could you explain this to me I don't get it. Like how I would find the element I know how to do electron configuration though

Answer 3

start from hydrogen (row 1)
go down to the 4th row (starts with potassium K)
group 5A is the 5th column (ignoring the transition metals) which is the column starting with nitrogen

Answer 4

this is kind of confusing but there are two numbering systems for groups

Answer 5

C?

Answer 6

Okay so I just googled it and the answer is a .... But, how is it the d value.

Answer 7

notice how Ars-enic comes after a full d block so you would expect d10

Answer 8

Isnt there an empty space going across the d block

Answer 9

but its empty... does it still count as d10

Answer 10

even if there aren't any elements

Answer 11

empty? the electrons from Sc to Zn are full

Answer 12

Oh okay I see.

Answer 13

yeah that should be it

the only other possibility would be 7A but let's try 17 for now

Answer 14

It worked

Answer 15

Sodium

Answer 16

that's a good guess but sodium is considered an alkali metal



the only nonmetal in the list is phosphorus

Answer 17

anything above the zigzag?

Answer 18

yes

Answer 19

Did I match it correctly

Answer 20

you need to switch P3+ and P3-
and Mg2+ and Mg2-

if there are more electrons than protons the overall charge will be negative

Answer 21

pay attention to the group # and position on the ptable. elements toward the left side generally make positive ions and elements towards the right side generally make negative ions

K is in group 1 so it gets a +1 charge
Mg and Ba are in group 2 so they get a 2+ charge
S and O are in group 6A so they need 2 electrons to get the full 8 valance electrons, so they'll take 2 extra electrons to get -2 charge
Al is always +3 (you just need to memorize this)
N needs 3 electrons --> -3 charge
I only needs 1 electron --> -1 charge

Answer 22

In the ground-state electron configuration of Fe3+, how many unpaired electrons are present?

Answer 23

4?

Answer 24

I got [Ar]4s^0 2d^5

Answer 25

so would it be 4 or 5 because the 4s there isn't any paired and then for 2d5 there is one that is unpaired?

Answer 26

each of the 5 electrons in the 2d subshell gets its own orbital, so 5 unpaired

Answer 27

oh right. but what about the 4s^0

Answer 28

would they still be paired even with the zero

Answer 29

4s^0 means no electrons are in the s orbital so they're not available for pairing

Answer 30

oh got it

Answer 31

Build the orbital diagram for the ion most likely formed by phosphorus.

Answer 32

I drew this

Answer 33

hm not quite

Phosphorus wants to attain the same electron config. as the next highest noble gas which is Argon

going from Neon to Argon we have

[Ne] 3s2 3p6

Answer 34

oh okay so the answer is [Ne]3s2 3p6

Answer 35

But wouldnt it be the noble gas before that?

Answer 36

like the one before p is Ne

Answer 37

right, so the previous noble gas is Ne so we'd use Ne instead of helium

Answer 38

oh right

Answer 39

I just inputted it wrong. I did orginally do 2p but i guess it didnt show up

Answer 40

the d-block should be empty as well

Answer 41

but it says an ion formed by phosphorous it would be the same right

Answer 42

so no 3d

Answer 43

right

Answer 44

it should be the same configuration as Argon

Answer 45

[Ar]3d^10

Answer 46

because it would condense down to Ar

Answer 47

it still needs to start with Neon

Answer 48

idk why its condensing down

Answer 49

so phosphorus wants to match the higher noble gas (Argon)

the configuration of Argon is 1s2 2s2 2p6 3s2 3p6 which is equivalent to [Ne] 3s2 3p6

Answer 50

Oh 3p^6 :S

Answer 51

why is the 2p orbital above the others

it should be between 2s and 3s

Answer 52

I got [Ne] 3s2 3p6 now

Answer 53

Is that right

Answer 54

the configuration is right, idk what's going on in your orbital diagram

Answer 55

Fixed it

Answer 56

Ugh it says its wrong:/

Answer 57

you need to switch 3s and 2p on the orbital diagram

if the configuration itself is wrong idk

Answer 58

wait so which comes first

Answer 59

2p?

Answer 60

1s2 2s2 2p6 3s2 3p6

2p comes before 3s

Answer 61

iirc phosphorus should be larger than sulfur since it's closer to the left side of the ptable

otherwise i think it's good

Answer 62

almost, for the last one Se is the bigger one (further down the group and further left on the ptable)

Fluorine will pretty much be smaller than everything

Answer 63

it says its wrong

Answer 64

You filled in 2 of 4 blanks incorrectly. Note that Sn and I both reside in period 5 of the periodic table (belonging to different groups). Consider the trend of atomic radius as you move right across a period.

Answer 65

this is what it says

Answer 66

huh

Answer 67

it syas its wrong and then gave me that^ explanation

Answer 68

oh duh, the further right you go the smaller the atoms are >_>

so Sn will be greater than I

Answer 69

and then cr is greater than w?

Answer 70

hm, no, W should be bigger

Answer 71

okay then one more that they are saying is wrong

Answer 72

oh ge?

Answer 73

is bigger?

Answer 74

just a hunch but they might want "not predictable" for Ge vs Po :S

Answer 75

okay so we got Sn , W, np, and f

Answer 76

it says one blank is wrong

Answer 77

Se is bigger than F

Answer 78

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
almost, for the last one Se is the bigger one (further down the group and further left on the ptable)

Fluorine will pretty much be smaller than everything
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 79

Got that one.. For this this is the radius size

Answer 80

So would it be the same concept

Answer 81

Rb,Ca,,S,Si,F

Answer 82

Thats what I put

Answer 83

Silicon is bigger than sulfur (it's closer to the left side)

Rb,Ca,,Si, S,F

Answer 84

C

Answer 85

hm, close but not quite, smaller atoms mean more charge, so Cl should have a larger ENC (effective nuclear charge) than P

Answer 86

a

Answer 87

So then a right

Answer 88

yeah that's what i got too

Answer 89

Ge,Cs,In,Se?

Answer 90

hm not quite, since it wants high to low ionization energy try going from the top right to the bottom left corner of the ptable

Answer 91

,In,Se, Ge, Cs?

Answer 92

almost

Se is the most towards the upper right

going straight left gives us Ge

then In, then Cs at the bottom

Answer 93

yikes

you need to draw out the orbital diagrams, and if there are unpaired electrons it's paramagnetic, if all the electrons are paired its diamagnetic

Answer 94

5 mins ahh

Answer 95

uh

going from right to left

F, O, N, B, Na, Li?

Answer 96

there may be some wonky stuff going on with O and N though

Answer 97

actually wait, switch the N and O, there's a full subshell on N which makes the energy higher

Answer 98

Which of the following trends is indirectly proportional to effective nuclear charge, Zeff?

Answer 99

ionic energy
welectron affinity
atomic size
all the answeres are correct?

Answer 100

atomic radii (just have to memorize this)

Answer 101

or atomic size i gues

Answer 102

Of Ca or Sr, the element with the higher first ionization energy is

Answer 103

Ca?

Answer 104

yes