Chemistry 101 - Introduction to the Periodic Table

CJam, 64 59 58 56 54 bytes

Thanks to Dennis for saving 2 bytes.

{_80-zG-z8<{80>"LA"=}{_"X8
"f>[-14_AAGH].*+:+Imd)}?}

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Leaves either period and group or a single character on the stack.

Explanation

There are two main ideas here:

• First, we deal with the Lanthanides and Actinides. We have the condition 56 < x < 72 for Lanthanides and 88 < x < 104 for Actinides. Both of these can be expressed as a single comparison by taking an absolute difference to the centre of the range: the inequalities become |x - 64| < 8 and |x - 96| < 8, respectively. But these are still very similar, and doing the two comparisons separately is expensive. So we apply the same idea of checking a symmetric range, by taking another absolute difference with the centre between the two ranges, 80, first: ||x-80| - 16| < 8. This condition indicates that the atom is either a Lanthanide or an Actinide, but distinguishing between these two cases is then as trivial as comparing against 80 (or some other value between the ranges).
• Since the output is really an index in a table of width 18, an obvious approach is to try base-converting the value to base 18, so that the two digits give group and period. To do that, we need to shift some values around though. All the we really need to do is to add the gaps in periods 1, 2 and 3 and close the gaps in periods 6 and 7. It's easiest to do this from the end, so that the values of the other gaps aren't affected (and keep their values).

_            e# Make a copy of the input, to figure out whether the output
             e# should be L or A.
80-z         e# Absolute difference from 80.
G-z          e# Absolute difference from 16.
8<           e# Check whether the result is less than 8.
{            e# If so...
  80>        e#   Check whether the input is greater than 80.
  "LA"=      e#   Select 'L or 'A accordingly.
}{           e# ...otherwise...
  _          e#   Duplicate input again.
  "X8"    e#   Push a string with character codes [88 56 12 4 1].
             e#   These are the values just before the gaps.
  f>         e#   Compare the input to each of these.
  [-14_AAGH] e#   Push [-14 -14 10 10 16 17]. These are the changes we need to
             e#   make to remove or insert the gaps corresponding to the above
             e#   positions. Note that the 17 doesn't get paired with an offset.
             e#   It's a constant offset to itself, which is equivalent to
             e#   decrementing the input (to make it zero based) and adding 18
             e#   to make the increment the period and make it 1-based.
  .*         e#   Multiply each gap by whether it's before the input value.
  +:+        e#   Add all of the applicable gaps to the input value.
  Imd        e#   Divmod 18, gives 1-based period and 0-based group.
  )          e#   Increment the group to make it one-based.
}?

05AB1E, 113 102 99 bytes

X18©XY‚Dˆ13®Ÿ¯13®Ÿ®LD¯15'L×S15L3+©¯15'A×S®)˜¹<è,XXY8×SD>4 18×SD>S66Sð14×S6 15×S77Sð15×S7 15×S)˜¹<è,

Explanation:

(start to construct the period part)
X18 ~ push 1 and 18
© ~ store 18 in register_c without p-opping
XY ~ push 1 and 2
13® ~ push 13 and the top element in register_c (18)
Ÿ ~ range - pop 2 values and push [a .. b]
XY ~ push 1 and 2
13®Ÿ ~ range - 13 to 18
XY ~ push 1, 2
13®Ÿ ~ range - 13 to 18
®LD ~ range - 1 to 18 (twice)
2L ~ range - 1 to 2
15'L×S ~ push 'L' 15 times
15L ~ range - 1 to 15
3+ ~ add 3 to each value in topmost element in stack
© ~ store in register-c without popping
2L ~ range - 1 to 2
15'A×S ~ push 'A' 15 times
® ~ push topmost value in register_c
) ~ wrap stack to array
˜ ~ deep flatten
¹<è ~ 1-indexed value of input in array
, ~ print & pop
(start to construct the group part)
XX ~ push 1 twice
Y8×SD ~ push 2 eight times (twice)
> ~ increment each value by 1
4 18×S ~ push 4 eighteen times (twice)
> ~ increment each value by one
66S ~ push 6 twice
ð15×S ~ push a space character 15 times
6 15×S ~ push 6 fifteen times
77S ~ push 7 two times
ð15×S ~ push a space character 15 times
7 15×S ~ push 7 fifteen times
)˜ ~ wrap stack to array and deep flatten
¹<è, ~ get 1-indexed value of input in array, then pop & print

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Mathematica, 77 bytes

e=ElementData;Which[56<#<72,"L",88<#<104,"A",1>0,{#~e~"Group",#~e~"Period"}]&

It would also be quite easy to use ElementData to determine whether the input is a Lanthanide or Actinide, but it would take about 20 more bytes.